3.1 Rechnungen mit komplexen Zahlen

Example 1

If we would like to find out the sum of the roots (solutions) of the equation x2−2x+2=0 we can first obtain the roots x1=1+−1  and x2=1−−1 . These roots contain the non-real number −1 . If we allow ourselves to do calculations containing −1  we see that the sum of x1 and x2 turns out to be 1+−1+1−−1=2 , which certainly is a real number.

In order to solve our problem we had to use a number that was not real in order to arrive at the real number solution.

Example 2

1. \displaystyle (3-5i)+(-4+i)=-1-4i
2. \displaystyle \bigl(\tfrac{1}{2}+2i\bigr)-\bigl(\tfrac{1}{6}+3i\bigr) = \tfrac{1}{3}-i
3. \displaystyle \frac{3+2i}{5}-\frac{3-i}{2} = \frac{6+4i}{10}-\frac{15-5i}{10} = \frac{-9+9i}{10} = -0\textrm{.}9 + 0\textrm{.}9i

Example 3

1. \displaystyle 3(4-i)=12-3i
2. \displaystyle 2i(3-5i)=6i-10i^2=10+6i
3. \displaystyle (1+i)(2+3i)=2+3i+2i+3i^2=-1+5i
4. \displaystyle (3+2i)(3-2i)=3^2-(2i)^2=9-4i^2=13
5. \displaystyle (3+i)^2=3^2+2\cdot3i+i^2=8+6i
6. \displaystyle i^{12}=(i^2)^6=(-1)^6=1
7. \displaystyle i^{23}=i^{22}\cdot i=(i^2)^{11}\cdot i=(-1)^{11}i=-i

Example 4

1. \displaystyle z=5+i\qquad then \displaystyle \quad\overline{z}=5-i\,.
2. \displaystyle z=-3-2i\qquad then \displaystyle \quad\overline{z} =-3+2i\,.
3. \displaystyle z=17\qquad then \displaystyle \quad\overline{z} =17\,.
4. \displaystyle z=i\qquad then \displaystyle \quad\overline{z} =-i\,.
5. \displaystyle z=-5i\qquad then \displaystyle \quad\overline{z} =5i\,.

Example 5

1. If \displaystyle z=4+3i one has
   • \displaystyle z+\overline{z} = 4 + 3i + 4 -3i = 8
   • \displaystyle z-\overline{z} = 6i
   • \displaystyle z \cdot \overline{z} = 4^2-(3i)^2=16+9=25
2. If for \displaystyle z one has \displaystyle \mathop{\rm Re} z=-2 and \displaystyle \mathop{\rm Im} z=1, one gets
   • \displaystyle z+\overline{z} = 2\,\mathop{\rm Re} z = -4
   • \displaystyle z-\overline{z} = 2i\,\mathop{\rm Im} z = 2i
   • \displaystyle z\cdot \overline{z} = (-2)^2+1^2=5

Example 6

1. \displaystyle \quad\frac{4+2i}{1+i} = \frac{(4+2i)(1-i)}{(1+i)(1-i)} = \frac{4-4i+2i-2i^2}{1-i^2} = \frac{6-2i}{2}=3-i
2. \displaystyle \quad\frac{25}{3-4i} = \frac{25(3+4i)}{(3-4i)(3+4i)} = \frac{25(3+4i)}{3^2-16i^2} = \frac{25(3+4i)}{25} = 3+4i
3. \displaystyle \quad\frac{3-2i}{i} = \frac{(3-2i)(-i)}{i(-i)} = \frac{-3i+2i^2}{-i^2} = \frac{-2-3i}{1} = -2-3i

Example 7

1. \displaystyle \quad\frac{2}{2-i}-\frac{i}{1+i} = \frac{2(2+i)}{(2-i)(2+i)} - \frac{i(1-i)}{(1+i)(1-i)} = \frac{4+2i}{5}-\frac{1+i}{2}
   \displaystyle \quad\phantom{\frac{2}{2-i}-\frac{i}{1+i}}{} = \frac{8+4i}{10}-\frac{5+5i}{10} = \frac{3-i}{10}\vphantom{\Biggl(}
2. \displaystyle \quad\frac{1-\dfrac{2}{1-i}}{2i+\dfrac{i}{2+i}} = \frac{\dfrac{1-i}{1-i}-\dfrac{2}{1-i}}{\dfrac{2i(2+i)}{(2+i)} + \dfrac{i}{2+i}} = \frac{\dfrac{1-i-2}{1-i}}{\dfrac{4i+2i^2 + i}{2+i}} = \frac{\dfrac{-1-i}{1-i}}{\dfrac{-2+5i}{2+i}}
   \displaystyle \quad\phantom{\smash{\frac{1-\dfrac{2}{1-i}}{2i+\dfrac{i}{2+i}}}}{} = \frac{-1-i}{1-i}\cdot \frac{2+i}{-2+5i} = \frac{(-1-i)(2+i)}{(1-i)(-2+5i)} = \frac{-2-i-2i-i^2}{-2+5i+2i-5i^2}\vphantom{\Biggl(}
   \displaystyle \quad\phantom{\smash{\frac{1-\dfrac{2}{1-i}}{2i+\dfrac{i}{2+i}}}}{} = \frac{-1-3i}{3+7i} = \frac{(-1-3i)(3-7i)}{(3+7i)(3-7i)} = \frac{-3+7i-9i+21i^2}{3^2-49i^2}\vphantom{\Biggl(} \displaystyle \quad\phantom{\smash{\frac{1-\dfrac{2}{1-i}}{2i+\dfrac{i}{2+i}}}}{} = \frac{-24-2i}{58} = \frac{-12-i}{29}\vphantom{\Biggl(}

Example 8

Determine the real number \displaystyle a so that the expression \displaystyle \ \frac{2-3i}{2+ai}\ becomes real.

Multiply the numerator and denominator with the complex conjugate of the denominator so that the expression can be written with separate real and imaginary parts.

1. REDIRECT Template:Abgesetzte Formel

If the expression is to be real , the imaginary part must be 0, ie.

1. REDIRECT Template:Abgesetzte Formel

Example 9

1. Solve the equation \displaystyle 3z+1-i=z-3+7i.
   
   Collect all \displaystyle z on the left-hand side by subtracting \displaystyle z from both sides
   1. REDIRECT Template:Abgesetzte Formel and now subtract \displaystyle 1-i
   2. REDIRECT Template:Abgesetzte Formel
   This gives that \displaystyle \ z=\frac{-4+8i}{2}=-2+4i\,\mbox{.}
2. Solve the equation \displaystyle z(-1-i)=6-2i.
   
   Divide both sides by \displaystyle -1-i in order to obtain \displaystyle z
   1. REDIRECT Template:Abgesetzte Formel
3. Solve the equation \displaystyle 3iz-2i=1-z.
   
   Adding \displaystyle z and \displaystyle 2i to both sides gives
   1. REDIRECT Template:Abgesetzte Formel
   This gives
   1. REDIRECT Template:Abgesetzte Formel
4. Solve the equation \displaystyle 2z+1-i=\bar z +3 + 2i.
   
   The equation contains both \displaystyle z as well as \displaystyle \overline{z} and therefore we assume \displaystyle z to be \displaystyle z=a+ib and solve the equation for \displaystyle a and \displaystyle b by equating the real and imaginary parts of both sides
   1. REDIRECT Template:Abgesetzte Formel
   i.e.
   1. REDIRECT Template:Abgesetzte Formel
   which gives
   1. REDIRECT Template:Abgesetzte Formel
   The answer is therefore, \displaystyle z=2+i.