3.3 Potenzen und Wurzeln

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The computational rules <math>\ \arg (zw) = \arg z + \arg w\ </math> and <math>\ |\,zw\,| = |\,z\,|\cdot|\,w\,|\ </math> mean that
The computational rules <math>\ \arg (zw) = \arg z + \arg w\ </math> and <math>\ |\,zw\,| = |\,z\,|\cdot|\,w\,|\ </math> mean that
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{{Fristående formel||<math>\biggl\{\begin{align*}&\arg (z\cdot z) = \arg z + \arg z \\ &|\,z\cdot z\,| = |\,z\,|\cdot|\,z\,|\end{align*}\qquad\biggl\{\begin{align*}&\arg z^3 = 3 \arg z \cr &|\,z^3\,| = |\,z\,|^3\end{align*}\qquad\text{etc.}</math>}}
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{{Displayed math||<math>\biggl\{\begin{align*}&\arg (z\cdot z) = \arg z + \arg z \\ &|\,z\cdot z\,| = |\,z\,|\cdot|\,z\,|\end{align*}\qquad\biggl\{\begin{align*}&\arg z^3 = 3 \arg z \cr &|\,z^3\,| = |\,z\,|^3\end{align*}\qquad\text{etc.}</math>}}
For an arbitrary number <math>z=r\,(\cos \alpha +i\,\sin \alpha)</math>, we therefore have the following relationship
For an arbitrary number <math>z=r\,(\cos \alpha +i\,\sin \alpha)</math>, we therefore have the following relationship
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{{Fristående formel||<math>z^n = \bigl(r\,(\cos \alpha +i\sin \alpha)\bigr)^n = r^n\,(\cos n\alpha +i\,\sin n\alpha)\,\mbox{.}</math>}}
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{{Displayed math||<math>z^n = \bigl(r\,(\cos \alpha +i\sin \alpha)\bigr)^n = r^n\,(\cos n\alpha +i\,\sin n\alpha)\,\mbox{.}</math>}}
If <math>|\,z\,|=1</math>, (i.e. <math>z</math> lies on the unit circle) then one has the special relationship
If <math>|\,z\,|=1</math>, (i.e. <math>z</math> lies on the unit circle) then one has the special relationship
<div class="regel">
<div class="regel">
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{{Fristående formel||<math>(\cos \alpha +i\,\sin \alpha)^n = \cos n\alpha +i\,\sin n\alpha\,\mbox{,}</math>}}
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{{Displayed math||<math>(\cos \alpha +i\,\sin \alpha)^n = \cos n\alpha +i\,\sin n\alpha\,\mbox{,}</math>}}
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</div>
Zeile 57: Zeile 57:
We write <math>z</math> in polar form <math>\ \ z= \frac{1}{\sqrt2} + \frac{i}{\sqrt2} = 1\cdot \Bigl(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\Bigr)\ \ </math> and Moivre's formula gives
We write <math>z</math> in polar form <math>\ \ z= \frac{1}{\sqrt2} + \frac{i}{\sqrt2} = 1\cdot \Bigl(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\Bigr)\ \ </math> and Moivre's formula gives
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{{Fristående formel||<math>\begin{align*}z^3 &= \Bigl( \cos\frac{\pi}{4} + i\,\sin\frac{\pi}{4}\,\Bigr)^3 = \cos\frac{3\pi}{4} + i\,\sin\frac{3\pi}{4} = -\frac{1}{\sqrt2} + \frac{1}{\sqrt2}\,i = \frac{-1+i}{\sqrt2}\,\mbox{,}\\[6pt] z^{100} &= \Bigl( \cos\frac{\pi}{4} + i\,\sin\frac{\pi}{4}\,\Bigr)^{100} = \cos\frac{100\pi}{4} + i\,\sin\frac{100\pi}{4}\\[4pt] &= \cos 25\pi + i\,\sin 25\pi = \cos \pi + i\,\sin \pi = -1\,\mbox{.}\end{align*}</math>}}
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{{Displayed math||<math>\begin{align*}z^3 &= \Bigl( \cos\frac{\pi}{4} + i\,\sin\frac{\pi}{4}\,\Bigr)^3 = \cos\frac{3\pi}{4} + i\,\sin\frac{3\pi}{4} = -\frac{1}{\sqrt2} + \frac{1}{\sqrt2}\,i = \frac{-1+i}{\sqrt2}\,\mbox{,}\\[6pt] z^{100} &= \Bigl( \cos\frac{\pi}{4} + i\,\sin\frac{\pi}{4}\,\Bigr)^{100} = \cos\frac{100\pi}{4} + i\,\sin\frac{100\pi}{4}\\[4pt] &= \cos 25\pi + i\,\sin 25\pi = \cos \pi + i\,\sin \pi = -1\,\mbox{.}\end{align*}</math>}}
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In the usual way one does an expansion by means of the squaring rules
In the usual way one does an expansion by means of the squaring rules
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{{Fristående formel||<math>\begin{align*} (\cos v + i\,\sin v)^2 &= \cos^2\!v + i^2 \sin^2\!v + 2i \sin v \cos v\\ &= \cos^2\!v - \sin^2\!v + 2i \sin v \cos v\end{align*}</math>}}
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{{Displayed math||<math>\begin{align*} (\cos v + i\,\sin v)^2 &= \cos^2\!v + i^2 \sin^2\!v + 2i \sin v \cos v\\ &= \cos^2\!v - \sin^2\!v + 2i \sin v \cos v\end{align*}</math>}}
and according to de Moivre's formula one gets
and according to de Moivre's formula one gets
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{{Fristående formel||<math>(\cos v + i \sin v)^2 = \cos 2v + i \sin 2v\,\mbox{.}</math>}}
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{{Displayed math||<math>(\cos v + i \sin v)^2 = \cos 2v + i \sin 2v\,\mbox{.}</math>}}
If one equates the real and imaginary parts of the two expressions one gets the well-known trigonometric formulas
If one equates the real and imaginary parts of the two expressions one gets the well-known trigonometric formulas
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{{Fristående formel||<math>\biggl\{\begin{align*}\cos 2v &= \cos^2\!v - \sin^2\!v\,\mbox{,}\\[2pt] \sin 2v&= 2 \sin v \cos v\,\mbox{.}\end{align*}</math>}}
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{{Displayed math||<math>\biggl\{\begin{align*}\cos 2v &= \cos^2\!v - \sin^2\!v\,\mbox{,}\\[2pt] \sin 2v&= 2 \sin v \cos v\,\mbox{.}\end{align*}</math>}}
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Then we get with de Moivre's formula
Then we get with de Moivre's formula
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{{Fristående formel||<math>\frac{(\sqrt3 + i)^{14}}{(1+i\sqrt3\,)^7(1+i)^{10}} = \frac{\displaystyle 2^{14}\Bigl(\cos\frac{14\pi}{6} + i\,\sin \frac{14\pi}{6}\,\Bigr)\vphantom{\biggl(}}{\displaystyle 2^7\Bigl(\cos \frac{7\pi}{3} + i\,\sin\frac{7\pi}{3}\,\Bigr) \cdot (\sqrt{2}\,)^{10}\Bigl(\cos\frac{10\pi}{4} + i\,\sin\frac{10\pi}{4}\,\Bigr)\vphantom{\biggl(}}</math>}}
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{{Displayed math||<math>\frac{(\sqrt3 + i)^{14}}{(1+i\sqrt3\,)^7(1+i)^{10}} = \frac{\displaystyle 2^{14}\Bigl(\cos\frac{14\pi}{6} + i\,\sin \frac{14\pi}{6}\,\Bigr)\vphantom{\biggl(}}{\displaystyle 2^7\Bigl(\cos \frac{7\pi}{3} + i\,\sin\frac{7\pi}{3}\,\Bigr) \cdot (\sqrt{2}\,)^{10}\Bigl(\cos\frac{10\pi}{4} + i\,\sin\frac{10\pi}{4}\,\Bigr)\vphantom{\biggl(}}</math>}}
and this expression can be simplified by performing multiplication and division in polar form
and this expression can be simplified by performing multiplication and division in polar form
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{{Fristående formel||<math>\begin{align*}\frac{\displaystyle 2^{14}\Bigl(\cos\frac{14\pi}{6} + i\,\sin\frac{14\pi}{6}\,\Bigr)\vphantom{\biggl(}} {\displaystyle 2^{12}\Bigl(\cos\frac{29\pi}{6} + i\,\sin\frac{29\pi}{6}\,\Bigr)\vphantom{\biggl(}} &= 2^2 \Bigl(\cos\Bigl( -\frac{15\pi}{6}\,\Bigr) + i\,\sin\Bigl( -\frac{15\pi}{6}\,\Bigr)\,\Bigr)\\[8pt] &= 4\Bigl(\cos \Bigl( -\frac{\pi}{2}\,\Bigr) + i\,\sin\Bigl( -\frac{\pi}{2}\,\Bigr)\,\Bigr) = -4i\,\mbox{.}\end{align*}</math>}}
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{{Displayed math||<math>\begin{align*}\frac{\displaystyle 2^{14}\Bigl(\cos\frac{14\pi}{6} + i\,\sin\frac{14\pi}{6}\,\Bigr)\vphantom{\biggl(}} {\displaystyle 2^{12}\Bigl(\cos\frac{29\pi}{6} + i\,\sin\frac{29\pi}{6}\,\Bigr)\vphantom{\biggl(}} &= 2^2 \Bigl(\cos\Bigl( -\frac{15\pi}{6}\,\Bigr) + i\,\sin\Bigl( -\frac{15\pi}{6}\,\Bigr)\,\Bigr)\\[8pt] &= 4\Bigl(\cos \Bigl( -\frac{\pi}{2}\,\Bigr) + i\,\sin\Bigl( -\frac{\pi}{2}\,\Bigr)\,\Bigr) = -4i\,\mbox{.}\end{align*}</math>}}
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Zeile 106: Zeile 106:
A complex number <math>z</math> is called the ''n''th root of the complex number <math>w</math> if
A complex number <math>z</math> is called the ''n''th root of the complex number <math>w</math> if
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<div class="regel">
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{{Fristående formel||<math>z^n= w \mbox{.}</math>}}
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{{Displayed math||<math>z^n= w \mbox{.}</math>}}
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Zeile 112: Zeile 112:
For a given number <math>w=|\,w\,|\,(\cos \theta + i\,\sin \theta)</math> one assumes that <math>z=r\,(\cos \alpha + i\, \sin \alpha)</math> and after insertion, the binomial equation becomes
For a given number <math>w=|\,w\,|\,(\cos \theta + i\,\sin \theta)</math> one assumes that <math>z=r\,(\cos \alpha + i\, \sin \alpha)</math> and after insertion, the binomial equation becomes
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{{Fristående formel||<math>r^{\,n}\,(\cos n\alpha + i \sin n\alpha) =|w|\,(\cos \theta + i \sin \theta)\,\mbox{,}</math>}}
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{{Displayed math||<math>r^{\,n}\,(\cos n\alpha + i \sin n\alpha) =|w|\,(\cos \theta + i \sin \theta)\,\mbox{,}</math>}}
where de Moivres formula has been used on the left-hand side. Equating moduli and arguments gives
where de Moivres formula has been used on the left-hand side. Equating moduli and arguments gives
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{{Fristående formel||<math>\biggl\{\begin{align*} r^{\,n} &= |w|\,\mbox{,}\\ n\alpha &= \theta + k\cdot 2\pi\,\mbox{.}\end{align*}</math>}}
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{{Displayed math||<math>\biggl\{\begin{align*} r^{\,n} &= |w|\,\mbox{,}\\ n\alpha &= \theta + k\cdot 2\pi\,\mbox{.}\end{align*}</math>}}
Note that we add multiples of <math>2\pi</math> to include all possible values of the argument that have the same direction as <math>\theta</math>. One gets
Note that we add multiples of <math>2\pi</math> to include all possible values of the argument that have the same direction as <math>\theta</math>. One gets
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{{Fristående formel||<math>\biggl\{\begin{align*} r &={\textstyle\sqrt[\scriptstyle n]{|w|}},\\ \alpha &= (\theta + 2k\pi)/n\,, \quad k=0, \pm 1, \pm 2, \ldots\end{align*}</math>}}
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{{Displayed math||<math>\biggl\{\begin{align*} r &={\textstyle\sqrt[\scriptstyle n]{|w|}},\\ \alpha &= (\theta + 2k\pi)/n\,, \quad k=0, \pm 1, \pm 2, \ldots\end{align*}</math>}}
This gives ''one'' value of <math>r</math>, but infinitely many values of <math>\alpha</math>. Despite this, there are not infinitely many solutions. From <math>k = 0</math> to <math>k = n - 1</math> one gets different arguments for <math>z</math> and thus different positions for <math>z</math> in the complex plane. For the other values of <math>k</math> due to the periodicity of the sine and cosine, one returns to these positions and therefore no new solutions are obtained. This reasoning shows that the equation <math>z^n=w</math> has exactly <math>n</math> roots.
This gives ''one'' value of <math>r</math>, but infinitely many values of <math>\alpha</math>. Despite this, there are not infinitely many solutions. From <math>k = 0</math> to <math>k = n - 1</math> one gets different arguments for <math>z</math> and thus different positions for <math>z</math> in the complex plane. For the other values of <math>k</math> due to the periodicity of the sine and cosine, one returns to these positions and therefore no new solutions are obtained. This reasoning shows that the equation <math>z^n=w</math> has exactly <math>n</math> roots.
Zeile 139: Zeile 139:
This turns the equation <math>\ z^4=16\,i\ </math> into
This turns the equation <math>\ z^4=16\,i\ </math> into
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{{Fristående formel||<math>r^4\,(\cos 4\alpha + i\,\sin 4\alpha) = 16\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\,\Bigr)\,\mbox{.}</math>}}
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{{Displayed math||<math>r^4\,(\cos 4\alpha + i\,\sin 4\alpha) = 16\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\,\Bigr)\,\mbox{.}</math>}}
Matching the moduli and arguments on both sides gives
Matching the moduli and arguments on both sides gives
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{{Fristående formel||<math>\biggl\{\begin{align*} r^4 &= 16,\\ 4\alpha &= \pi/2 + k\cdot 2\pi,\end{align*}\qquad\text{i.e.}\qquad\biggl\{\begin{align*} r &= \sqrt[\scriptstyle 4]{16}= 2, \\ \alpha &= \pi/8 + k\pi/2\,,\quad k=0,1,2,3.\end{align*}</math>}}
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{{Displayed math||<math>\biggl\{\begin{align*} r^4 &= 16,\\ 4\alpha &= \pi/2 + k\cdot 2\pi,\end{align*}\qquad\text{i.e.}\qquad\biggl\{\begin{align*} r &= \sqrt[\scriptstyle 4]{16}= 2, \\ \alpha &= \pi/8 + k\pi/2\,,\quad k=0,1,2,3.\end{align*}</math>}}
{| width="100%"
{| width="100%"
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The solutions to the equation are thus
The solutions to the equation are thus
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{{Fristående formel||<math>\left\{\begin{align*}\displaystyle z_1&= 2\Bigl(\cos \frac{\pi}{8} + i\,\sin\frac{\pi}{8}\,\Bigr),\\[4pt]
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{{Displayed math||<math>\left\{\begin{align*}\displaystyle z_1&= 2\Bigl(\cos \frac{\pi}{8} + i\,\sin\frac{\pi}{8}\,\Bigr),\\[4pt]
\displaystyle z_2 &= 2\Bigl(\cos\frac{5\pi}{8} + i\,\sin\frac{5\pi}{8}\,\Bigr),\vphantom{\biggl(}\\[4pt]
\displaystyle z_2 &= 2\Bigl(\cos\frac{5\pi}{8} + i\,\sin\frac{5\pi}{8}\,\Bigr),\vphantom{\biggl(}\\[4pt]
\displaystyle z_3 &= 2\Bigl(\cos\frac{9\pi}{8} + i\,\sin\frac{9\pi}{8}\,\Bigr),\vphantom{\biggl(}\\[4pt]
\displaystyle z_3 &= 2\Bigl(\cos\frac{9\pi}{8} + i\,\sin\frac{9\pi}{8}\,\Bigr),\vphantom{\biggl(}\\[4pt]
Zeile 162: Zeile 162:
If we manipulate <math>i</math> as if it were a real number and treat a complex number <math>z</math> as a function of just <math>\alpha</math> ( where <math>r</math> is a constant),
If we manipulate <math>i</math> as if it were a real number and treat a complex number <math>z</math> as a function of just <math>\alpha</math> ( where <math>r</math> is a constant),
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{{Fristående formel||<math>f(\alpha) = r\,(\cos \alpha + i\,\sin \alpha)</math>}}
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{{Displayed math||<math>f(\alpha) = r\,(\cos \alpha + i\,\sin \alpha)</math>}}
we get after differentiation
we get after differentiation
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{{Fristående formel||<math>\begin{align*} f^{\,\prime}(\alpha) &= -r\sin \alpha + r\,i\,\cos \alpha = r\,i^2 \sin \alpha + r\,i\,\cos \alpha = i\,r\,(\cos \alpha + i\,\sin \alpha) = i\,f(\alpha)\\ f^{\,\prime\prime} (\alpha) &= - r\,\cos \alpha - r\,i\,\sin \alpha = i^2\,r\,(\cos \alpha + i\,\sin \alpha) = i^2\, f(\alpha)\cr &\text{etc.}\end{align*}</math>}}
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{{Displayed math||<math>\begin{align*} f^{\,\prime}(\alpha) &= -r\sin \alpha + r\,i\,\cos \alpha = r\,i^2 \sin \alpha + r\,i\,\cos \alpha = i\,r\,(\cos \alpha + i\,\sin \alpha) = i\,f(\alpha)\\ f^{\,\prime\prime} (\alpha) &= - r\,\cos \alpha - r\,i\,\sin \alpha = i^2\,r\,(\cos \alpha + i\,\sin \alpha) = i^2\, f(\alpha)\cr &\text{etc.}\end{align*}</math>}}
The only real-valued functions which behave like this are <math>f(x)= e^{\,kx}</math>, which justifies the definition
The only real-valued functions which behave like this are <math>f(x)= e^{\,kx}</math>, which justifies the definition
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{{Fristående formel||<math>e^{\,i\alpha} = \cos \alpha + i\,\sin \alpha\,\mbox{.}</math>}}
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{{Displayed math||<math>e^{\,i\alpha} = \cos \alpha + i\,\sin \alpha\,\mbox{.}</math>}}
This definition turns out to be a completely natural generalisation of the exponential function for the real numbers. Putting <math>z=a+ib</math> one gets
This definition turns out to be a completely natural generalisation of the exponential function for the real numbers. Putting <math>z=a+ib</math> one gets
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{{Fristående formel||<math>e^{\,z} = e^{\,a+ib} = e^{\,a} \cdot e^{\,ib} = e^{\,a}(\cos b + i\,\sin b)\,\mbox{.}</math>}}
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{{Displayed math||<math>e^{\,z} = e^{\,a+ib} = e^{\,a} \cdot e^{\,ib} = e^{\,a}(\cos b + i\,\sin b)\,\mbox{.}</math>}}
The definition of <math>e^{\,z}</math> may be regarded as a convenient notation for the polar form of a complex number, as <math>z=r\,(\cos \alpha + i\,\sin \alpha) = r\,e^{\,i\alpha}\,</math>.
The definition of <math>e^{\,z}</math> may be regarded as a convenient notation for the polar form of a complex number, as <math>z=r\,(\cos \alpha + i\,\sin \alpha) = r\,e^{\,i\alpha}\,</math>.
Zeile 185: Zeile 185:
For a real number <math>z</math> the definition is consistent with the case when the exponent is real, as <math>z=a+0\cdot i</math> which gives
For a real number <math>z</math> the definition is consistent with the case when the exponent is real, as <math>z=a+0\cdot i</math> which gives
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{{Fristående formel||<math>e^{\,z} = e^{\,a+0\cdot i} = e^a (\cos 0 + i \sin 0) = e^a \cdot 1 = e^a\,\mbox{.}</math>}}
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{{Displayed math||<math>e^{\,z} = e^{\,a+0\cdot i} = e^a (\cos 0 + i \sin 0) = e^a \cdot 1 = e^a\,\mbox{.}</math>}}
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Zeile 194: Zeile 194:
A further indication of why the above definition is so natural is given by the relationship
A further indication of why the above definition is so natural is given by the relationship
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{{Fristående formel||<math>\bigl(e^{\,i\alpha}\bigr)^n = (\cos \alpha + i \sin \alpha)^n = \cos n\alpha + i \sin n \alpha = e^{\,in\alpha}\,\mbox{,}</math>}}
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{{Displayed math||<math>\bigl(e^{\,i\alpha}\bigr)^n = (\cos \alpha + i \sin \alpha)^n = \cos n\alpha + i \sin n \alpha = e^{\,in\alpha}\,\mbox{,}</math>}}
which demonstrates that de Moivre's formula is actually identical to the well-known law of exponents,
which demonstrates that de Moivre's formula is actually identical to the well-known law of exponents,
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{{Fristående formel||<math>\left(a^x\right)^y = a^{x\,y}\,\mbox{.}</math>}}
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{{Displayed math||<math>\left(a^x\right)^y = a^{x\,y}\,\mbox{.}</math>}}
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Zeile 208: Zeile 208:
From the above definition, one can obtain the relationship
From the above definition, one can obtain the relationship
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{{Fristående formel||<math>e^{\pi\,i} = \cos \pi + i \sin \pi = -1</math>}}
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{{Displayed math||<math>e^{\pi\,i} = \cos \pi + i \sin \pi = -1</math>}}
which connects together the, generally regarded, most basic numbers in mathematics: <math>e</math>, <math>\pi</math>, <math>i</math> and 1.
which connects together the, generally regarded, most basic numbers in mathematics: <math>e</math>, <math>\pi</math>, <math>i</math> and 1.
Zeile 227: Zeile 227:
The equation in polar form is <math>\ r^3e^{3\alpha i}=8\,e^{3\pi i/2}\ </math> and matching the moduli and arguments on both sides gives,
The equation in polar form is <math>\ r^3e^{3\alpha i}=8\,e^{3\pi i/2}\ </math> and matching the moduli and arguments on both sides gives,
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{{Fristående formel||<math>\biggl\{\begin{align*} r^3 &= 8\,\mbox{,}\\ 3\alpha &= 3\pi/2+2k\pi\,\mbox{,}\end{align*}\qquad\Leftrightarrow\qquad\biggl\{\begin{align*} r&=\sqrt[\scriptstyle 3]{8}\,\mbox{,}\\ \alpha&= \pi/2+2k\pi/3\,,\quad k=0,1,2\,\mbox{.}\end{align*}</math>}}
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{{Displayed math||<math>\biggl\{\begin{align*} r^3 &= 8\,\mbox{,}\\ 3\alpha &= 3\pi/2+2k\pi\,\mbox{,}\end{align*}\qquad\Leftrightarrow\qquad\biggl\{\begin{align*} r&=\sqrt[\scriptstyle 3]{8}\,\mbox{,}\\ \alpha&= \pi/2+2k\pi/3\,,\quad k=0,1,2\,\mbox{.}\end{align*}</math>}}
The roots of the equation are thus
The roots of the equation are thus
Zeile 247: Zeile 247:
If for <math>z=a+ib</math> one has <math>|\,z\,|=r</math> and <math>\arg z = \alpha</math> then for <math>\overline{z}= a-ib</math> one gets <math>|\,\overline{z}\,|=r</math> and <math>\arg \overline{z} = - \alpha</math>.This means that <math>z=r\,e^{i\alpha}</math> and <math>\overline{z} = r\,e^{-i\alpha}</math>. The equation can be written
If for <math>z=a+ib</math> one has <math>|\,z\,|=r</math> and <math>\arg z = \alpha</math> then for <math>\overline{z}= a-ib</math> one gets <math>|\,\overline{z}\,|=r</math> and <math>\arg \overline{z} = - \alpha</math>.This means that <math>z=r\,e^{i\alpha}</math> and <math>\overline{z} = r\,e^{-i\alpha}</math>. The equation can be written
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{{Fristående formel||<math>(r\,e^{i\alpha})^2 = r\,e^{-i\alpha}\qquad\text{or}\qquad r^2 e^{2i\alpha}= r\,e^{-i\alpha}\,\mbox{,}</math>}}
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{{Displayed math||<math>(r\,e^{i\alpha})^2 = r\,e^{-i\alpha}\qquad\text{or}\qquad r^2 e^{2i\alpha}= r\,e^{-i\alpha}\,\mbox{,}</math>}}
which directly gives that <math>r=0</math> is a solution, i.e. <math>z=0</math>. If we assume that <math>r\not=0</math> then the equation can be written as <math>\ r\,e^{3i\alpha} = 1\,</math>, which gives after matching moduli and arguments
which directly gives that <math>r=0</math> is a solution, i.e. <math>z=0</math>. If we assume that <math>r\not=0</math> then the equation can be written as <math>\ r\,e^{3i\alpha} = 1\,</math>, which gives after matching moduli and arguments
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{{Fristående formel||<math>\biggl\{\begin{align*} r &= 1\,\mbox{,}\\ 3\alpha &= 0 + 2k\pi\,\mbox{,}\end{align*}\qquad\Leftrightarrow\qquad\biggl\{\begin{align*} r &= 1\,\mbox{,}\\ \alpha &= 2k\pi/3\,\mbox{,}\quad k=0,1,2\,\mbox{.}\end{align*}</math>}}
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{{Displayed math||<math>\biggl\{\begin{align*} r &= 1\,\mbox{,}\\ 3\alpha &= 0 + 2k\pi\,\mbox{,}\end{align*}\qquad\Leftrightarrow\qquad\biggl\{\begin{align*} r &= 1\,\mbox{,}\\ \alpha &= 2k\pi/3\,\mbox{,}\quad k=0,1,2\,\mbox{.}\end{align*}</math>}}
The solutions are
The solutions are
Zeile 266: Zeile 266:
The squaring rules,
The squaring rules,
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{{Fristående formel||<math>\left\{\begin{align*} (a+b)^2 &= a^2+2ab+b^2\\ (a-b)^2 &= a^2-2ab+b^2\end{align*}\right.</math>}}
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{{Displayed math||<math>\left\{\begin{align*} (a+b)^2 &= a^2+2ab+b^2\\ (a-b)^2 &= a^2-2ab+b^2\end{align*}\right.</math>}}
which are usually used to expand parenthesis can also be used in reverse to obtain quadratic expressions. For example,
which are usually used to expand parenthesis can also be used in reverse to obtain quadratic expressions. For example,
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{{Fristående formel||<math>\begin{align*} x^2+4x+4 &= (x+2)^2\,\mbox{,}\\ x^2-10x+25 &= (x-5)^2\,\mbox{.}\end{align*}</math>}}
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{{Displayed math||<math>\begin{align*} x^2+4x+4 &= (x+2)^2\,\mbox{,}\\ x^2-10x+25 &= (x-5)^2\,\mbox{.}\end{align*}</math>}}
This can be used to solve quadratic equations, for example,
This can be used to solve quadratic equations, for example,
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{{Fristående formel||<math>\begin{align*} x^2+4x+4 &= 9\,\mbox{,}\\ (x+2)^2 &= 9\,\mbox{.}\end{align*}</math>}}
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{{Displayed math||<math>\begin{align*} x^2+4x+4 &= 9\,\mbox{,}\\ (x+2)^2 &= 9\,\mbox{.}\end{align*}</math>}}
Taking roots then gives that <math>x+2=\pm\sqrt{9}</math> and thus that <math>x=-2\pm 3</math>, i.e. <math>x=1</math> or <math>x=-5</math>.
Taking roots then gives that <math>x+2=\pm\sqrt{9}</math> and thus that <math>x=-2\pm 3</math>, i.e. <math>x=1</math> or <math>x=-5</math>.
Zeile 281: Zeile 281:
Sometimes it is necessary to add or subtract an appropriate number to obtain a suitable expression. The above equation, for example, could just as easily been presented to us as
Sometimes it is necessary to add or subtract an appropriate number to obtain a suitable expression. The above equation, for example, could just as easily been presented to us as
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{{Fristående formel||<math>x^2+4x-5=0\,\mbox{.}</math>}}
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{{Displayed math||<math>x^2+4x-5=0\,\mbox{.}</math>}}
By adding 9 to both sides, we get a suitable expression on the left side:
By adding 9 to both sides, we get a suitable expression on the left side:
-
{{Fristående formel||<math>\begin{align*} x^2+4x-5+9 &= 0+9\,\mbox{,}\\ x^2+4x+4\phantom{{}+9} &= 9\,\mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*} x^2+4x-5+9 &= 0+9\,\mbox{,}\\ x^2+4x+4\phantom{{}+9} &= 9\,\mbox{.}\end{align*}</math>}}
This method is called ''completing the square''.
This method is called ''completing the square''.
Zeile 299: Zeile 299:
The coefficient in front of <math>x</math> is <math>-6</math> and it shows that we must have the number <math>(-3)^2=9</math> as the constant term on the left-hand side to make a complete square. By adding <math>2</math> to both sides we achieve this:
The coefficient in front of <math>x</math> is <math>-6</math> and it shows that we must have the number <math>(-3)^2=9</math> as the constant term on the left-hand side to make a complete square. By adding <math>2</math> to both sides we achieve this:
-
{{Fristående formel||<math>\begin{align*} x^2-6x+7+2 &= 2+2\,\mbox{,}\\ x^2-6x+9\phantom{{}+2} &= 4\,\mbox{,}\\ \rlap{(x-3)^2}\phantom{x^2-6x+7+2}{} &= 4\,\mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*} x^2-6x+7+2 &= 2+2\,\mbox{,}\\ x^2-6x+9\phantom{{}+2} &= 4\,\mbox{,}\\ \rlap{(x-3)^2}\phantom{x^2-6x+7+2}{} &= 4\,\mbox{.}\end{align*}</math>}}
Taking roots then gives <math>x-3=\pm 2</math>, which means that <math>x=1</math> or <math>x=5</math>.
Taking roots then gives <math>x-3=\pm 2</math>, which means that <math>x=1</math> or <math>x=5</math>.
Zeile 309: Zeile 309:
The equation can be written as <math>z^2+8z+17=0</math>. By subtracting 1 on both sides, we get a complete square on the left-hand side:
The equation can be written as <math>z^2+8z+17=0</math>. By subtracting 1 on both sides, we get a complete square on the left-hand side:
-
{{Fristående formel||<math>\begin{align*} z^2+8z+17-1 &= 0-1\,\mbox{,}\\ z^2+8z+16\phantom{{}-1} &= -1\,\mbox{,}\\ \rlap{(z+4)^2}\phantom{z^2+8z+17-1}{} &= -1\,\mbox{,}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*} z^2+8z+17-1 &= 0-1\,\mbox{,}\\ z^2+8z+16\phantom{{}-1} &= -1\,\mbox{,}\\ \rlap{(z+4)^2}\phantom{z^2+8z+17-1}{} &= -1\,\mbox{,}\end{align*}</math>}}
and thus <math>z+4=\pm\sqrt{-1}</math>. In other words, the solutions are <math>z=-4-i</math> and <math>z=-4+i</math>.
and thus <math>z+4=\pm\sqrt{-1}</math>. In other words, the solutions are <math>z=-4-i</math> and <math>z=-4+i</math>.
Zeile 329: Zeile 329:
Half the coefficient of <math>x</math> is <math>-\tfrac{4}{3}</math>. We thus add <math>\bigl(-\tfrac{4}{3}\bigr)^2=\tfrac{16}{9}</math> to both sides
Half the coefficient of <math>x</math> is <math>-\tfrac{4}{3}</math>. We thus add <math>\bigl(-\tfrac{4}{3}\bigr)^2=\tfrac{16}{9}</math> to both sides
-
{{Fristående formel||<math>\begin{align*} x^2-\tfrac{8}{3}x+\tfrac{16}{9}+1 &= 2+\tfrac{16}{9}\,\mbox{,}\\ \rlap{\bigl(x-\tfrac{4}{3}\bigr)^2}\phantom{x^2-\tfrac{8}{3}x+\tfrac{16}{9}}{}+1 &= \tfrac{34}{9}\,\mbox{,}\\ \rlap{\bigl(x-\tfrac{4}{3}\bigr)^2}\phantom{x^2-\tfrac{8}{3}x+\tfrac{16}{9}+1} &= \tfrac{25}{9}\,\mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*} x^2-\tfrac{8}{3}x+\tfrac{16}{9}+1 &= 2+\tfrac{16}{9}\,\mbox{,}\\ \rlap{\bigl(x-\tfrac{4}{3}\bigr)^2}\phantom{x^2-\tfrac{8}{3}x+\tfrac{16}{9}}{}+1 &= \tfrac{34}{9}\,\mbox{,}\\ \rlap{\bigl(x-\tfrac{4}{3}\bigr)^2}\phantom{x^2-\tfrac{8}{3}x+\tfrac{16}{9}+1} &= \tfrac{25}{9}\,\mbox{.}\end{align*}</math>}}
Now it's easy to get to <math>x-\tfrac{4}{3}=\pm\tfrac{5}{3}</math> and thus to get that <math>x=\tfrac{4}{3}\pm\tfrac{5}{3}</math>, i.e. <math>x=-\tfrac{1}{3}</math> or <math>x=3</math>.
Now it's easy to get to <math>x-\tfrac{4}{3}=\pm\tfrac{5}{3}</math> and thus to get that <math>x=\tfrac{4}{3}\pm\tfrac{5}{3}</math>, i.e. <math>x=-\tfrac{1}{3}</math> or <math>x=3</math>.
Zeile 344: Zeile 344:
Completing the square gives
Completing the square gives
-
{{Fristående formel||<math>\begin{align*} x^2+px+\Bigl(\frac{p}{2}\Bigr)^2+q &= \Bigl(\frac{p}{2}\Bigr)^2\,\mbox{,}\\ \rlap{\Bigl(x+\frac{p}{2}\Bigr)^2}\phantom{x^2+px+\Bigl(\frac{p}{2}\Bigr)^2+q}{} &= \Bigl(\frac{p}{2}\Bigr)^2-q\,\mbox{,}\\ \rlap{x+\frac{p}{2}}\phantom{x^2+px+\Bigl(\frac{p}{2}\Bigr)^2+q}{} &= \pm \sqrt{\Bigl(\frac{p}{2}\Bigr)^2-q}\ \mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*} x^2+px+\Bigl(\frac{p}{2}\Bigr)^2+q &= \Bigl(\frac{p}{2}\Bigr)^2\,\mbox{,}\\ \rlap{\Bigl(x+\frac{p}{2}\Bigr)^2}\phantom{x^2+px+\Bigl(\frac{p}{2}\Bigr)^2+q}{} &= \Bigl(\frac{p}{2}\Bigr)^2-q\,\mbox{,}\\ \rlap{x+\frac{p}{2}}\phantom{x^2+px+\Bigl(\frac{p}{2}\Bigr)^2+q}{} &= \pm \sqrt{\Bigl(\frac{p}{2}\Bigr)^2-q}\ \mbox{.}\end{align*}</math>}}
This gives the usual formula, ''pq-formula'', for solutions to quadratic equations
This gives the usual formula, ''pq-formula'', for solutions to quadratic equations
-
{{Fristående formel||<math>x=-\frac{p}{2}\pm \sqrt{\Bigl(\frac{p}{2}\Bigr)^2-q}\,\mbox{.}</math>}}
+
{{Displayed math||<math>x=-\frac{p}{2}\pm \sqrt{\Bigl(\frac{p}{2}\Bigr)^2-q}\,\mbox{.}</math>}}
</div>
</div>
Zeile 362: Zeile 362:
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{{Fristående formel||<math>z^2-(12+4i)z+(-(6+2i))^2-4+24i=(-(6+2i))^2\,\mbox{.}</math>}}
+
{{Displayed math||<math>z^2-(12+4i)z+(-(6+2i))^2-4+24i=(-(6+2i))^2\,\mbox{.}</math>}}
Expanding the square on the right-hand side <math>\ (-(6+2i))^2=36+24i+4i^2=32+24i\ </math> and completing the square on the left-hand side gives
Expanding the square on the right-hand side <math>\ (-(6+2i))^2=36+24i+4i^2=32+24i\ </math> and completing the square on the left-hand side gives
-
{{Fristående formel||<math>\begin{align*} (z-(6+2i))^2-4+24i &= 32+24i\,\mbox{,}\\ \rlap{(z-(6+2i))^2}\phantom{(z-(6+2i))^2-4+24i}{} &= 36\,\mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*} (z-(6+2i))^2-4+24i &= 32+24i\,\mbox{,}\\ \rlap{(z-(6+2i))^2}\phantom{(z-(6+2i))^2-4+24i}{} &= 36\,\mbox{.}\end{align*}</math>}}
After a taking roots, we have that <math>\ z-(6+2i)=\pm 6\ </math> and therefore the solutions are <math>z=12+2i</math> and <math>z=2i</math>.
After a taking roots, we have that <math>\ z-(6+2i)=\pm 6\ </math> and therefore the solutions are <math>z=12+2i</math> and <math>z=2i</math>.
Zeile 374: Zeile 374:
If one wants to bring about a square in an expression one can use the same technique. In order not to change the value of the expression one both adds and subtracts the missing constant term, such as in the following,
If one wants to bring about a square in an expression one can use the same technique. In order not to change the value of the expression one both adds and subtracts the missing constant term, such as in the following,
-
{{Fristående formel||<math>\begin{align*} x^2+10x+3 &= x^2+10x+25+3-25\\ &= (x+5)^2-22\,\mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*} x^2+10x+3 &= x^2+10x+25+3-25\\ &= (x+5)^2-22\,\mbox{.}\end{align*}</math>}}
Zeile 386: Zeile 386:
Add and subtract the term <math>\bigl(\frac{1}{2}(2-4i)\bigr)^2=(1-2i)^2=-3-4i\,</math>,
Add and subtract the term <math>\bigl(\frac{1}{2}(2-4i)\bigr)^2=(1-2i)^2=-3-4i\,</math>,
-
{{Fristående formel||<math>\begin{align*} z^2+(2-4i)z+1-3i &= z^2+(2-4i)z+(1-2i)^2-(1-2i)^2+1-3i\\ &= \bigl(z+(1-2i)\bigr)^2-(1-2i)^2+1-3i\\ &= \bigl(z+(1-2i)\bigr)^2-(-3-4i)+1-3i\\ &= \bigl(z+(1-2i)\bigr)^2+4+i\,\mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*} z^2+(2-4i)z+1-3i &= z^2+(2-4i)z+(1-2i)^2-(1-2i)^2+1-3i\\ &= \bigl(z+(1-2i)\bigr)^2-(1-2i)^2+1-3i\\ &= \bigl(z+(1-2i)\bigr)^2-(-3-4i)+1-3i\\ &= \bigl(z+(1-2i)\bigr)^2+4+i\,\mbox{.}\end{align*}</math>}}
</div>
</div>
Zeile 395: Zeile 395:
To solve quadratic equations sometimes the simplest method is to use the usual formula for quadratic equations. However, this may lead to that one ends up with terms of the type <math>\sqrt{a+ib}</math>. One can then assume
To solve quadratic equations sometimes the simplest method is to use the usual formula for quadratic equations. However, this may lead to that one ends up with terms of the type <math>\sqrt{a+ib}</math>. One can then assume
-
{{Fristående formel||<math>z=x+iy=\sqrt{a+ib}\,\mbox{.}</math>}}
+
{{Displayed math||<math>z=x+iy=\sqrt{a+ib}\,\mbox{.}</math>}}
By squaring both sides we get
By squaring both sides we get
-
{{Fristående formel||<math>\begin{align*} (x+iy)^2 &= a+ib\,\mbox{,}\\ x^2 - y^2 + 2xy\,i &= a+ib\,\mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*} (x+iy)^2 &= a+ib\,\mbox{,}\\ x^2 - y^2 + 2xy\,i &= a+ib\,\mbox{.}\end{align*}</math>}}
Matching the real and imaginary parts gives
Matching the real and imaginary parts gives
-
{{Fristående formel||<math>\left\{\begin{align*} &x^2 - y^2 = a\,\mbox{,}\\ &2xy=b\,\mbox{.}\end{align*}\right.</math>}}
+
{{Displayed math||<math>\left\{\begin{align*} &x^2 - y^2 = a\,\mbox{,}\\ &2xy=b\,\mbox{.}\end{align*}\right.</math>}}
These equations can be solved by substitution, for example, <math>y= b/(2x)</math> can be inserted in the first equation.
These equations can be solved by substitution, for example, <math>y= b/(2x)</math> can be inserted in the first equation.
Zeile 417: Zeile 417:
Put <math>\ x+iy=\sqrt{-3-4i}\ </math> where <math>x</math> and <math>y</math> are real numbers. Squaring both sides gives
Put <math>\ x+iy=\sqrt{-3-4i}\ </math> where <math>x</math> and <math>y</math> are real numbers. Squaring both sides gives
-
{{Fristående formel||<math>\begin{align*} (x+iy)^2 &= -3-4i\,\mbox{,}\\ x^2 - y^2 + 2xyi &= -3-4i\,\mbox{,}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*} (x+iy)^2 &= -3-4i\,\mbox{,}\\ x^2 - y^2 + 2xyi &= -3-4i\,\mbox{,}\end{align*}</math>}}
which leads to the system of equations
which leads to the system of equations
-
{{Fristående formel||<math>\Bigl\{\begin{align*} x^2 - y^2 &= -3\,\mbox{,}\\ 2xy&= -4\,\mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>\Bigl\{\begin{align*} x^2 - y^2 &= -3\,\mbox{,}\\ 2xy&= -4\,\mbox{.}\end{align*}</math>}}
From the second equation, we can solve for <math>\ y=-4/(2x) = -2/x\ </math> and put it into the first equation to get
From the second equation, we can solve for <math>\ y=-4/(2x) = -2/x\ </math> and put it into the first equation to get
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{{Fristående formel||<math>x^2-\frac{4}{x^2} = -3 \quad \Leftrightarrow \quad x^4 +3x^2 - 4=0\,\mbox{.}</math>}}
+
{{Displayed math||<math>x^2-\frac{4}{x^2} = -3 \quad \Leftrightarrow \quad x^4 +3x^2 - 4=0\,\mbox{.}</math>}}
This is a quadratic equation in <math>x^2</math> which can be seen more easily by putting <math>t=x^2</math>,
This is a quadratic equation in <math>x^2</math> which can be seen more easily by putting <math>t=x^2</math>,
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{{Fristående formel||<math>t^2 +3t -4=0\,\mbox{.}</math>}}
+
{{Displayed math||<math>t^2 +3t -4=0\,\mbox{.}</math>}}
The solutions are <math>t = 1</math> and <math>t = -4</math>. The latter solution must be rejected, as <math>x</math> and <math>y</math> have been assumed to be real numbers, and thus <math>x^2=-4</math> cannot be true. We get <math>x=\pm\sqrt{1}</math>, which gives us two possible solutions
The solutions are <math>t = 1</math> and <math>t = -4</math>. The latter solution must be rejected, as <math>x</math> and <math>y</math> have been assumed to be real numbers, and thus <math>x^2=-4</math> cannot be true. We get <math>x=\pm\sqrt{1}</math>, which gives us two possible solutions
Zeile 437: Zeile 437:
So we can conclude that
So we can conclude that
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{{Fristående formel||<math>\sqrt{-3-4i} = \biggl\{\begin{align*} &\phantom{-}1-2i\,\mbox{,}\\ &-1+2i\,\mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>\sqrt{-3-4i} = \biggl\{\begin{align*} &\phantom{-}1-2i\,\mbox{,}\\ &-1+2i\,\mbox{.}\end{align*}</math>}}
</div>
</div>
Zeile 451: Zeile 451:
The formula for solutions to a quadratic equations (see example 3) gives that
The formula for solutions to a quadratic equations (see example 3) gives that
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{{Fristående formel||<math>z= 1\pm \sqrt{1-10} = 1\pm \sqrt{-9}= 1\pm 3i\,\mbox{.}</math>}}
+
{{Displayed math||<math>z= 1\pm \sqrt{1-10} = 1\pm \sqrt{-9}= 1\pm 3i\,\mbox{.}</math>}}
</li>
</li>
Zeile 458: Zeile 458:
<br>
<br>
Here, once again , the ''pq''-formula may be used giving the solutions directly
Here, once again , the ''pq''-formula may be used giving the solutions directly
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{{Fristående formel||<math>\begin{align*} z &= -2+i\pm\sqrt{\smash{(-2+i)^2+4i}\vphantom{i^2}} = -2+i\pm\sqrt{4-4i+i^{\,2}+4i}\\ &=-2+i\pm\sqrt{3} = -2\pm\sqrt{3}+i\,\mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*} z &= -2+i\pm\sqrt{\smash{(-2+i)^2+4i}\vphantom{i^2}} = -2+i\pm\sqrt{4-4i+i^{\,2}+4i}\\ &=-2+i\pm\sqrt{3} = -2\pm\sqrt{3}+i\,\mbox{.}\end{align*}</math>}}
</li>
</li>
Zeile 466: Zeile 466:
Division of both sides with <math>i</math> gives
Division of both sides with <math>i</math> gives
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{{Fristående formel||<math>\begin{align*} z^2 + \frac{2+6i}{i}z +\frac{2+11i}{i} &= 0\,\mbox{,}\\ z^2+ (6-2i)z + 11-2i &= 0\,\mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*} z^2 + \frac{2+6i}{i}z +\frac{2+11i}{i} &= 0\,\mbox{,}\\ z^2+ (6-2i)z + 11-2i &= 0\,\mbox{.}\end{align*}</math>}}
Applying the ''pq''- formula gives
Applying the ''pq''- formula gives
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{{Fristående formel||<math>\begin{align*} z &= -3+i \pm \sqrt{\smash{(-3+i)^2 -(11-2i)}\vphantom{i^2}}\\ &= -3+i \pm \sqrt{-3-4i}\\ &= -3+i\pm(1-2i)\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*} z &= -3+i \pm \sqrt{\smash{(-3+i)^2 -(11-2i)}\vphantom{i^2}}\\ &= -3+i \pm \sqrt{-3-4i}\\ &= -3+i\pm(1-2i)\end{align*}</math>}}
where we used the resulting value of<math>\ \sqrt{-3-4i}\ </math> which we obtained in example 15. The solutions are therefore
where we used the resulting value of<math>\ \sqrt{-3-4i}\ </math> which we obtained in example 15. The solutions are therefore
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{{Fristående formel||<math>z=\biggl\{\begin{align*} &-2-i\,\mbox{,}\\ &-4+3i\,\mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>z=\biggl\{\begin{align*} &-2-i\,\mbox{,}\\ &-4+3i\,\mbox{.}\end{align*}</math>}}
</li>
</li>
</ol>
</ol>
</div>
</div>

Version vom 08:25, 17. Sep. 2008

       Theory          Exercises      

Contents:

  • De Moivre's formula
  • Binomial equations
  • Exponential function
  • Euler's formula
  • Completing the square
  • Quadratic equations

Learning outcomes:

After this section, you will have learned how to:

  • Calculate the powers of complex numbers with de Moivre's formula.
  • Calculate the roots of certain complex numbers by rewriting to polar form.
  • Solve binomial equations.
  • Complete the square for complex quadratic expressions.
  • Solve complex quadratic equations.

De Moivre's formula

The computational rules \displaystyle \ \arg (zw) = \arg z + \arg w\ and \displaystyle \ |\,zw\,| = |\,z\,|\cdot|\,w\,|\ mean that

\displaystyle \biggl\{\begin{align*}&\arg (z\cdot z) = \arg z + \arg z \\ &|\,z\cdot z\,| = |\,z\,|\cdot|\,z\,|\end{align*}\qquad\biggl\{\begin{align*}&\arg z^3 = 3 \arg z \cr &|\,z^3\,| = |\,z\,|^3\end{align*}\qquad\text{etc.}

For an arbitrary number \displaystyle z=r\,(\cos \alpha +i\,\sin \alpha), we therefore have the following relationship

\displaystyle z^n = \bigl(r\,(\cos \alpha +i\sin \alpha)\bigr)^n = r^n\,(\cos n\alpha +i\,\sin n\alpha)\,\mbox{.}

If \displaystyle |\,z\,|=1, (i.e. \displaystyle z lies on the unit circle) then one has the special relationship

\displaystyle (\cos \alpha +i\,\sin \alpha)^n = \cos n\alpha +i\,\sin n\alpha\,\mbox{,}

which is usually referred to as de Moivres formula. This relationship is very useful when it comes to deriving trigonometric identities and calculating the roots and powers of complex numbers.


Example 1


If \displaystyle z = \frac{1+i}{\sqrt2}, determine \displaystyle z^3 and \displaystyle z^{100}.


We write \displaystyle z in polar form \displaystyle \ \ z= \frac{1}{\sqrt2} + \frac{i}{\sqrt2} = 1\cdot \Bigl(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\Bigr)\ \ and Moivre's formula gives

\displaystyle \begin{align*}z^3 &= \Bigl( \cos\frac{\pi}{4} + i\,\sin\frac{\pi}{4}\,\Bigr)^3 = \cos\frac{3\pi}{4} + i\,\sin\frac{3\pi}{4} = -\frac{1}{\sqrt2} + \frac{1}{\sqrt2}\,i = \frac{-1+i}{\sqrt2}\,\mbox{,}\\[6pt] z^{100} &= \Bigl( \cos\frac{\pi}{4} + i\,\sin\frac{\pi}{4}\,\Bigr)^{100} = \cos\frac{100\pi}{4} + i\,\sin\frac{100\pi}{4}\\[4pt] &= \cos 25\pi + i\,\sin 25\pi = \cos \pi + i\,\sin \pi = -1\,\mbox{.}\end{align*}

Example 2


In the usual way one does an expansion by means of the squaring rules

\displaystyle \begin{align*} (\cos v + i\,\sin v)^2 &= \cos^2\!v + i^2 \sin^2\!v + 2i \sin v \cos v\\ &= \cos^2\!v - \sin^2\!v + 2i \sin v \cos v\end{align*}

and according to de Moivre's formula one gets

\displaystyle (\cos v + i \sin v)^2 = \cos 2v + i \sin 2v\,\mbox{.}

If one equates the real and imaginary parts of the two expressions one gets the well-known trigonometric formulas


\displaystyle \biggl\{\begin{align*}\cos 2v &= \cos^2\!v - \sin^2\!v\,\mbox{,}\\[2pt] \sin 2v&= 2 \sin v \cos v\,\mbox{.}\end{align*}

Example 3


Simplify \displaystyle \ \ \frac{(\sqrt3 + i)^{14}}{(1+i\sqrt3\,)^7(1+i)^{10}}\,.


We write the numbers \displaystyle \sqrt{3}+i, \displaystyle 1+i\sqrt{3} and \displaystyle 1+i in polar form

  • \displaystyle \quad\sqrt{3} + i = 2\Bigl(\cos\frac{\pi}{6} + i\,\sin\frac{\pi}{6}\,\Bigr)\vphantom{\biggl(},
  • \displaystyle \quad 1+i\sqrt{3} = 2\Bigl(\cos\frac{\pi}{3} + i\,\sin\frac{\pi}{3}\,\Bigr)\vphantom{\biggl(},
  • \displaystyle \quad 1+i = \sqrt2\,\Bigl(\cos\frac{\pi}{4} + i\,\sin\frac{\pi}{4}\,\Bigr)\vphantom{\biggl(}.

Then we get with de Moivre's formula

\displaystyle \frac{(\sqrt3 + i)^{14}}{(1+i\sqrt3\,)^7(1+i)^{10}} = \frac{\displaystyle 2^{14}\Bigl(\cos\frac{14\pi}{6} + i\,\sin \frac{14\pi}{6}\,\Bigr)\vphantom{\biggl(}}{\displaystyle 2^7\Bigl(\cos \frac{7\pi}{3} + i\,\sin\frac{7\pi}{3}\,\Bigr) \cdot (\sqrt{2}\,)^{10}\Bigl(\cos\frac{10\pi}{4} + i\,\sin\frac{10\pi}{4}\,\Bigr)\vphantom{\biggl(}}

and this expression can be simplified by performing multiplication and division in polar form

\displaystyle \begin{align*}\frac{\displaystyle 2^{14}\Bigl(\cos\frac{14\pi}{6} + i\,\sin\frac{14\pi}{6}\,\Bigr)\vphantom{\biggl(}} {\displaystyle 2^{12}\Bigl(\cos\frac{29\pi}{6} + i\,\sin\frac{29\pi}{6}\,\Bigr)\vphantom{\biggl(}} &= 2^2 \Bigl(\cos\Bigl( -\frac{15\pi}{6}\,\Bigr) + i\,\sin\Bigl( -\frac{15\pi}{6}\,\Bigr)\,\Bigr)\\[8pt] &= 4\Bigl(\cos \Bigl( -\frac{\pi}{2}\,\Bigr) + i\,\sin\Bigl( -\frac{\pi}{2}\,\Bigr)\,\Bigr) = -4i\,\mbox{.}\end{align*}


Binomial equations

A complex number \displaystyle z is called the nth root of the complex number \displaystyle w if

\displaystyle z^n= w \mbox{.}

The above relationship can also be seen as an equation in which \displaystyle z is unknown. This type of equation is called a binomial equation. The solutions are obtained by rewriting both sides in polar form and comparing both the moduli and the arguments.

For a given number \displaystyle w=|\,w\,|\,(\cos \theta + i\,\sin \theta) one assumes that \displaystyle z=r\,(\cos \alpha + i\, \sin \alpha) and after insertion, the binomial equation becomes

\displaystyle r^{\,n}\,(\cos n\alpha + i \sin n\alpha) =|w|\,(\cos \theta + i \sin \theta)\,\mbox{,}

where de Moivres formula has been used on the left-hand side. Equating moduli and arguments gives

\displaystyle \biggl\{\begin{align*} r^{\,n} &= |w|\,\mbox{,}\\ n\alpha &= \theta + k\cdot 2\pi\,\mbox{.}\end{align*}

Note that we add multiples of \displaystyle 2\pi to include all possible values of the argument that have the same direction as \displaystyle \theta. One gets

\displaystyle \biggl\{\begin{align*} r &={\textstyle\sqrt[\scriptstyle n]{|w|}},\\ \alpha &= (\theta + 2k\pi)/n\,, \quad k=0, \pm 1, \pm 2, \ldots\end{align*}

This gives one value of \displaystyle r, but infinitely many values of \displaystyle \alpha. Despite this, there are not infinitely many solutions. From \displaystyle k = 0 to \displaystyle k = n - 1 one gets different arguments for \displaystyle z and thus different positions for \displaystyle z in the complex plane. For the other values of \displaystyle k due to the periodicity of the sine and cosine, one returns to these positions and therefore no new solutions are obtained. This reasoning shows that the equation \displaystyle z^n=w has exactly \displaystyle n roots.

Comment. Note that the arguments of the roots differ from each other by \displaystyle 2\pi/n so that the roots are evenly distributed on a circle with radius \displaystyle \sqrt[\scriptstyle n]{|w|} and form corners in a regular n-gon (an n sided polygon).


Exempel 4


Solve the binomial equation \displaystyle \ z^4= 16\,i\,.


Write \displaystyle z and \displaystyle 16\,i in polar form

  • \displaystyle \quad z=r\,(\cos \alpha + i\,\sin \alpha)\,,
  • \displaystyle \quad 16\,i= 16\Bigl(\cos\frac{\pi}{2} + i\,\sin\frac{\pi}{2}\,\Bigr)\vphantom{\biggl(}.

This turns the equation \displaystyle \ z^4=16\,i\ into

\displaystyle r^4\,(\cos 4\alpha + i\,\sin 4\alpha) = 16\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\,\Bigr)\,\mbox{.}

Matching the moduli and arguments on both sides gives

\displaystyle \biggl\{\begin{align*} r^4 &= 16,\\ 4\alpha &= \pi/2 + k\cdot 2\pi,\end{align*}\qquad\text{i.e.}\qquad\biggl\{\begin{align*} r &= \sqrt[\scriptstyle 4]{16}= 2, \\ \alpha &= \pi/8 + k\pi/2\,,\quad k=0,1,2,3.\end{align*}

The solutions to the equation are thus

\displaystyle \left\{\begin{align*}\displaystyle z_1&= 2\Bigl(\cos \frac{\pi}{8} + i\,\sin\frac{\pi}{8}\,\Bigr),\\[4pt]

\displaystyle z_2 &= 2\Bigl(\cos\frac{5\pi}{8} + i\,\sin\frac{5\pi}{8}\,\Bigr),\vphantom{\biggl(}\\[4pt] \displaystyle z_3 &= 2\Bigl(\cos\frac{9\pi}{8} + i\,\sin\frac{9\pi}{8}\,\Bigr),\vphantom{\biggl(}\\[4pt] \displaystyle z_4 &= 2\Bigl(\cos\frac{13\pi}{8} + i\,\sin\frac{13\pi}{8}\,\Bigr).\end{align*}\right.

3.3 - Figur - Komplexa talen z₁, z₂, z₃ och z₄


Exponential form of complex numbers

If we manipulate \displaystyle i as if it were a real number and treat a complex number \displaystyle z as a function of just \displaystyle \alpha ( where \displaystyle r is a constant),

\displaystyle f(\alpha) = r\,(\cos \alpha + i\,\sin \alpha)

we get after differentiation

\displaystyle \begin{align*} f^{\,\prime}(\alpha) &= -r\sin \alpha + r\,i\,\cos \alpha = r\,i^2 \sin \alpha + r\,i\,\cos \alpha = i\,r\,(\cos \alpha + i\,\sin \alpha) = i\,f(\alpha)\\ f^{\,\prime\prime} (\alpha) &= - r\,\cos \alpha - r\,i\,\sin \alpha = i^2\,r\,(\cos \alpha + i\,\sin \alpha) = i^2\, f(\alpha)\cr &\text{etc.}\end{align*}

The only real-valued functions which behave like this are \displaystyle f(x)= e^{\,kx}, which justifies the definition

\displaystyle e^{\,i\alpha} = \cos \alpha + i\,\sin \alpha\,\mbox{.}

This definition turns out to be a completely natural generalisation of the exponential function for the real numbers. Putting \displaystyle z=a+ib one gets

\displaystyle e^{\,z} = e^{\,a+ib} = e^{\,a} \cdot e^{\,ib} = e^{\,a}(\cos b + i\,\sin b)\,\mbox{.}

The definition of \displaystyle e^{\,z} may be regarded as a convenient notation for the polar form of a complex number, as \displaystyle z=r\,(\cos \alpha + i\,\sin \alpha) = r\,e^{\,i\alpha}\,.


Example 5


For a real number \displaystyle z the definition is consistent with the case when the exponent is real, as \displaystyle z=a+0\cdot i which gives

\displaystyle e^{\,z} = e^{\,a+0\cdot i} = e^a (\cos 0 + i \sin 0) = e^a \cdot 1 = e^a\,\mbox{.}

Example 6


A further indication of why the above definition is so natural is given by the relationship

\displaystyle \bigl(e^{\,i\alpha}\bigr)^n = (\cos \alpha + i \sin \alpha)^n = \cos n\alpha + i \sin n \alpha = e^{\,in\alpha}\,\mbox{,}

which demonstrates that de Moivre's formula is actually identical to the well-known law of exponents,

\displaystyle \left(a^x\right)^y = a^{x\,y}\,\mbox{.}

Example 7


From the above definition, one can obtain the relationship

\displaystyle e^{\pi\,i} = \cos \pi + i \sin \pi = -1

which connects together the, generally regarded, most basic numbers in mathematics: \displaystyle e, \displaystyle \pi, \displaystyle i and 1. This relationship is seen by many as the most beautiful in mathematics and was discovered by Euler in the early 1700's.

Example 8


Solve the equation \displaystyle \ (z+i)^3 = -8i.


Put \displaystyle w = z + i. We then get the binomial equation \displaystyle \ w^3=-8i\,. To begin with, we rewrite \displaystyle w and \displaystyle -8i in polar form

  • \displaystyle \quad w=r\,(\cos \alpha + i\,\sin \alpha) = r\,e^{i\alpha}\,\mbox{,}
  • \displaystyle \quad -8i = 8\Bigl(\cos \frac{3\pi}{2} + i\,\sin\frac{3\pi}{2}\,\Bigr) = 8\,e^{3\pi i/2}\vphantom{\biggl(}\,\mbox{.}

The equation in polar form is \displaystyle \ r^3e^{3\alpha i}=8\,e^{3\pi i/2}\ and matching the moduli and arguments on both sides gives,

\displaystyle \biggl\{\begin{align*} r^3 &= 8\,\mbox{,}\\ 3\alpha &= 3\pi/2+2k\pi\,\mbox{,}\end{align*}\qquad\Leftrightarrow\qquad\biggl\{\begin{align*} r&=\sqrt[\scriptstyle 3]{8}\,\mbox{,}\\ \alpha&= \pi/2+2k\pi/3\,,\quad k=0,1,2\,\mbox{.}\end{align*}

The roots of the equation are thus

  • \displaystyle \quad w_1 = 2\,e^{\pi i/2} = 2\Bigl(\cos \frac{\pi}{2} + i\,\sin\frac{\pi}{2}\,\Bigr) = 2i\,\mbox{,}\quad\vphantom{\biggl(}
  • \displaystyle \quad w_2 = 2\,e^{7\pi i/6} = 2\Bigl(\cos\frac{7\pi}{6} + i\,\sin\frac{7\pi}{6}\,\Bigr) = -\sqrt{3}-i\,\mbox{,}\quad\vphantom{\Biggl(}
  • \displaystyle \quad w_3 = 2\,e^{11\pi i/6} = 2\Bigl(\cos\frac{11\pi}{6} + i\,\sin\frac{11\pi}{6}\,\Bigr) = \sqrt{3}-i\,\mbox{,}\quad\vphantom{\biggl(}

i.e. \displaystyle z_1 = 2i-i=i, \displaystyle z_2 = - \sqrt{3}-2i and \displaystyle z_3 = \sqrt{3}-2i.

Example 9


Solve \displaystyle \ z^2 = \overline{z}\,.


If for \displaystyle z=a+ib one has \displaystyle |\,z\,|=r and \displaystyle \arg z = \alpha then for \displaystyle \overline{z}= a-ib one gets \displaystyle |\,\overline{z}\,|=r and \displaystyle \arg \overline{z} = - \alpha.This means that \displaystyle z=r\,e^{i\alpha} and \displaystyle \overline{z} = r\,e^{-i\alpha}. The equation can be written

\displaystyle (r\,e^{i\alpha})^2 = r\,e^{-i\alpha}\qquad\text{or}\qquad r^2 e^{2i\alpha}= r\,e^{-i\alpha}\,\mbox{,}

which directly gives that \displaystyle r=0 is a solution, i.e. \displaystyle z=0. If we assume that \displaystyle r\not=0 then the equation can be written as \displaystyle \ r\,e^{3i\alpha} = 1\,, which gives after matching moduli and arguments

\displaystyle \biggl\{\begin{align*} r &= 1\,\mbox{,}\\ 3\alpha &= 0 + 2k\pi\,\mbox{,}\end{align*}\qquad\Leftrightarrow\qquad\biggl\{\begin{align*} r &= 1\,\mbox{,}\\ \alpha &= 2k\pi/3\,\mbox{,}\quad k=0,1,2\,\mbox{.}\end{align*}

The solutions are

  • \displaystyle \quad z_1 = e^0 = 1\,\mbox{,}
  • \displaystyle \quad z_2 = e^{2\pi i/ 3} = \cos\frac{2\pi}{3} + i\,\sin\frac{2\pi}{3} = -\frac{1}{2} + \frac{\sqrt3}{2}\,i\,\mbox{,}\vphantom{\Biggl(}
  • \displaystyle \quad z_3 = e^{4\pi i/ 3} = \cos\frac{4\pi}{3} + i\,\sin\frac{4\pi}{3} = -\frac{1}{2} - \frac{\sqrt3}{2}\,i\,\mbox{,}
  • \displaystyle \quad z_4 = 0\,\mbox{.}


Completing the square

The squaring rules,

\displaystyle \left\{\begin{align*} (a+b)^2 &= a^2+2ab+b^2\\ (a-b)^2 &= a^2-2ab+b^2\end{align*}\right.

which are usually used to expand parenthesis can also be used in reverse to obtain quadratic expressions. For example,

\displaystyle \begin{align*} x^2+4x+4 &= (x+2)^2\,\mbox{,}\\ x^2-10x+25 &= (x-5)^2\,\mbox{.}\end{align*}

This can be used to solve quadratic equations, for example,

\displaystyle \begin{align*} x^2+4x+4 &= 9\,\mbox{,}\\ (x+2)^2 &= 9\,\mbox{.}\end{align*}

Taking roots then gives that \displaystyle x+2=\pm\sqrt{9} and thus that \displaystyle x=-2\pm 3, i.e. \displaystyle x=1 or \displaystyle x=-5.


Sometimes it is necessary to add or subtract an appropriate number to obtain a suitable expression. The above equation, for example, could just as easily been presented to us as

\displaystyle x^2+4x-5=0\,\mbox{.}

By adding 9 to both sides, we get a suitable expression on the left side:

\displaystyle \begin{align*} x^2+4x-5+9 &= 0+9\,\mbox{,}\\ x^2+4x+4\phantom{{}+9} &= 9\,\mbox{.}\end{align*}

This method is called completing the square.


Example 10

  1. Solve the equation \displaystyle \ x^2-6x+7=2\,.

    The coefficient in front of \displaystyle x is \displaystyle -6 and it shows that we must have the number \displaystyle (-3)^2=9 as the constant term on the left-hand side to make a complete square. By adding \displaystyle 2 to both sides we achieve this:
    \displaystyle \begin{align*} x^2-6x+7+2 &= 2+2\,\mbox{,}\\ x^2-6x+9\phantom{{}+2} &= 4\,\mbox{,}\\ \rlap{(x-3)^2}\phantom{x^2-6x+7+2}{} &= 4\,\mbox{.}\end{align*}

    Taking roots then gives \displaystyle x-3=\pm 2, which means that \displaystyle x=1 or \displaystyle x=5.

  2. Solve the equation \displaystyle \ z^2+21=4-8z\,.

    The equation can be written as \displaystyle z^2+8z+17=0. By subtracting 1 on both sides, we get a complete square on the left-hand side:
    \displaystyle \begin{align*} z^2+8z+17-1 &= 0-1\,\mbox{,}\\ z^2+8z+16\phantom{{}-1} &= -1\,\mbox{,}\\ \rlap{(z+4)^2}\phantom{z^2+8z+17-1}{} &= -1\,\mbox{,}\end{align*}

    and thus \displaystyle z+4=\pm\sqrt{-1}. In other words, the solutions are \displaystyle z=-4-i and \displaystyle z=-4+i.

Generally, completing the square may be regarded as arranging that "the square of half the coefficient of the x-term" is the constant term in the quadratic expression. This term can always be added to the two sides without worrying about the other terms and then manipulating the equation. If the coefficients of the expression are complex numbers, one still can go about it in the same way.


Example 11


Solve the equation \displaystyle \ x^2-\frac{8}{3}x+1=2\,.


Half the coefficient of \displaystyle x is \displaystyle -\tfrac{4}{3}. We thus add \displaystyle \bigl(-\tfrac{4}{3}\bigr)^2=\tfrac{16}{9} to both sides

\displaystyle \begin{align*} x^2-\tfrac{8}{3}x+\tfrac{16}{9}+1 &= 2+\tfrac{16}{9}\,\mbox{,}\\ \rlap{\bigl(x-\tfrac{4}{3}\bigr)^2}\phantom{x^2-\tfrac{8}{3}x+\tfrac{16}{9}}{}+1 &= \tfrac{34}{9}\,\mbox{,}\\ \rlap{\bigl(x-\tfrac{4}{3}\bigr)^2}\phantom{x^2-\tfrac{8}{3}x+\tfrac{16}{9}+1} &= \tfrac{25}{9}\,\mbox{.}\end{align*}

Now it's easy to get to \displaystyle x-\tfrac{4}{3}=\pm\tfrac{5}{3} and thus to get that \displaystyle x=\tfrac{4}{3}\pm\tfrac{5}{3}, i.e. \displaystyle x=-\tfrac{1}{3} or \displaystyle x=3.

Example 12


Solve the equation \displaystyle \ x^2+px+q=0\,.


Completing the square gives

\displaystyle \begin{align*} x^2+px+\Bigl(\frac{p}{2}\Bigr)^2+q &= \Bigl(\frac{p}{2}\Bigr)^2\,\mbox{,}\\ \rlap{\Bigl(x+\frac{p}{2}\Bigr)^2}\phantom{x^2+px+\Bigl(\frac{p}{2}\Bigr)^2+q}{} &= \Bigl(\frac{p}{2}\Bigr)^2-q\,\mbox{,}\\ \rlap{x+\frac{p}{2}}\phantom{x^2+px+\Bigl(\frac{p}{2}\Bigr)^2+q}{} &= \pm \sqrt{\Bigl(\frac{p}{2}\Bigr)^2-q}\ \mbox{.}\end{align*}

This gives the usual formula, pq-formula, for solutions to quadratic equations

\displaystyle x=-\frac{p}{2}\pm \sqrt{\Bigl(\frac{p}{2}\Bigr)^2-q}\,\mbox{.}

Example 13


Solve the equation \displaystyle \ z^2-(12+4i)z-4+24i=0\,.


Half the coefficient of \displaystyle z is \displaystyle -(6+2i) so we add the square of this expression to both sides


\displaystyle z^2-(12+4i)z+(-(6+2i))^2-4+24i=(-(6+2i))^2\,\mbox{.}

Expanding the square on the right-hand side \displaystyle \ (-(6+2i))^2=36+24i+4i^2=32+24i\ and completing the square on the left-hand side gives

\displaystyle \begin{align*} (z-(6+2i))^2-4+24i &= 32+24i\,\mbox{,}\\ \rlap{(z-(6+2i))^2}\phantom{(z-(6+2i))^2-4+24i}{} &= 36\,\mbox{.}\end{align*}

After a taking roots, we have that \displaystyle \ z-(6+2i)=\pm 6\ and therefore the solutions are \displaystyle z=12+2i and \displaystyle z=2i.

If one wants to bring about a square in an expression one can use the same technique. In order not to change the value of the expression one both adds and subtracts the missing constant term, such as in the following,

\displaystyle \begin{align*} x^2+10x+3 &= x^2+10x+25+3-25\\ &= (x+5)^2-22\,\mbox{.}\end{align*}


Example 14


Complete the square in the expression \displaystyle \ z^2+(2-4i)z+1-3i\,.


Add and subtract the term \displaystyle \bigl(\frac{1}{2}(2-4i)\bigr)^2=(1-2i)^2=-3-4i\,,

\displaystyle \begin{align*} z^2+(2-4i)z+1-3i &= z^2+(2-4i)z+(1-2i)^2-(1-2i)^2+1-3i\\ &= \bigl(z+(1-2i)\bigr)^2-(1-2i)^2+1-3i\\ &= \bigl(z+(1-2i)\bigr)^2-(-3-4i)+1-3i\\ &= \bigl(z+(1-2i)\bigr)^2+4+i\,\mbox{.}\end{align*}


Solving using a formula

To solve quadratic equations sometimes the simplest method is to use the usual formula for quadratic equations. However, this may lead to that one ends up with terms of the type \displaystyle \sqrt{a+ib}. One can then assume

\displaystyle z=x+iy=\sqrt{a+ib}\,\mbox{.}

By squaring both sides we get

\displaystyle \begin{align*} (x+iy)^2 &= a+ib\,\mbox{,}\\ x^2 - y^2 + 2xy\,i &= a+ib\,\mbox{.}\end{align*}

Matching the real and imaginary parts gives

\displaystyle \left\{\begin{align*} &x^2 - y^2 = a\,\mbox{,}\\ &2xy=b\,\mbox{.}\end{align*}\right.

These equations can be solved by substitution, for example, \displaystyle y= b/(2x) can be inserted in the first equation.


Example 15


Calculate \displaystyle \ \sqrt{-3-4i}\,.


Put \displaystyle \ x+iy=\sqrt{-3-4i}\ where \displaystyle x and \displaystyle y are real numbers. Squaring both sides gives

\displaystyle \begin{align*} (x+iy)^2 &= -3-4i\,\mbox{,}\\ x^2 - y^2 + 2xyi &= -3-4i\,\mbox{,}\end{align*}

which leads to the system of equations

\displaystyle \Bigl\{\begin{align*} x^2 - y^2 &= -3\,\mbox{,}\\ 2xy&= -4\,\mbox{.}\end{align*}

From the second equation, we can solve for \displaystyle \ y=-4/(2x) = -2/x\ and put it into the first equation to get

\displaystyle x^2-\frac{4}{x^2} = -3 \quad \Leftrightarrow \quad x^4 +3x^2 - 4=0\,\mbox{.}

This is a quadratic equation in \displaystyle x^2 which can be seen more easily by putting \displaystyle t=x^2,

\displaystyle t^2 +3t -4=0\,\mbox{.}

The solutions are \displaystyle t = 1 and \displaystyle t = -4. The latter solution must be rejected, as \displaystyle x and \displaystyle y have been assumed to be real numbers, and thus \displaystyle x^2=-4 cannot be true. We get \displaystyle x=\pm\sqrt{1}, which gives us two possible solutions

  • \displaystyle \ x=-1\ which gives \displaystyle \ y=-2/(-1)=2\,,
  • \displaystyle \ x=1\ which gives \displaystyle \ y=-2/1=-2\,.

So we can conclude that

\displaystyle \sqrt{-3-4i} = \biggl\{\begin{align*} &\phantom{-}1-2i\,\mbox{,}\\ &-1+2i\,\mbox{.}\end{align*}

Example 16


  1. Solve the equation \displaystyle \ z^2-2z+10=0\,.

    The formula for solutions to a quadratic equations (see example 3) gives that
    \displaystyle z= 1\pm \sqrt{1-10} = 1\pm \sqrt{-9}= 1\pm 3i\,\mbox{.}
  2. Solve the equation \displaystyle \ z^2 + (4-2i)z -4i=0\,\mbox{.}

    Here, once again , the pq-formula may be used giving the solutions directly
    \displaystyle \begin{align*} z &= -2+i\pm\sqrt{\smash{(-2+i)^2+4i}\vphantom{i^2}} = -2+i\pm\sqrt{4-4i+i^{\,2}+4i}\\ &=-2+i\pm\sqrt{3} = -2\pm\sqrt{3}+i\,\mbox{.}\end{align*}
  3. Solve the equation \displaystyle \ iz^2+(2+6i)z+2+11i=0\,\mbox{.}

    Division of both sides with \displaystyle i gives
    \displaystyle \begin{align*} z^2 + \frac{2+6i}{i}z +\frac{2+11i}{i} &= 0\,\mbox{,}\\ z^2+ (6-2i)z + 11-2i &= 0\,\mbox{.}\end{align*}

    Applying the pq- formula gives

    \displaystyle \begin{align*} z &= -3+i \pm \sqrt{\smash{(-3+i)^2 -(11-2i)}\vphantom{i^2}}\\ &= -3+i \pm \sqrt{-3-4i}\\ &= -3+i\pm(1-2i)\end{align*}

    where we used the resulting value of\displaystyle \ \sqrt{-3-4i}\ which we obtained in example 15. The solutions are therefore

    \displaystyle z=\biggl\{\begin{align*} &-2-i\,\mbox{,}\\ &-4+3i\,\mbox{.}\end{align*}
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