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Aus Online Mathematik Brückenkurs 2

As usual we begin by completing the square,

z−21+3i2−21+3i2−4+3iz−21+3i2−41+23i+49i2−4+3iz−21+3i2−41−23i+49−4+3iz−21+3i2−2+23i=0=0=0=0

and if we treat w=z−21+3i as unknown, we have the equation

w=2−23i.

Up until now, we have solved binomial equations of this type by going over to polar form, but if we were to do that in this case, we would have a problem with determining the exact value of the argument of the right-hand side. Instead, we put w=x+iy and try to obtain x and y from the equation.

With w=x+iy, the equation becomes

(x+iy)2=2−23i

and, with the left-hand side expanded,

x2−y2+2xyi=2−23i.

If we set the real and imaginary part of both sides equal, we obtain the equation system

x2−y22xy=2=−23.

We would very well be able to solve this system, but there is a further relation that we can obtain which will simplify the calculations. If we go back to the relation

(x+iy)2=2−23i

and take the modulus of both sides, we obtain

x2+y2=22+−232

i.e.

x2+y2=25.

We add this relation to our two other equations,

x2−y22xyx2+y2=2=−23=25.

Now, add the first and third equations

	x2	−	y2	=	2
+	x2	+	y2	=	25
	2x2			=	29

which gives that x=23. Then, subtracting the first equation from the third,

	x2	+	y2	\displaystyle {}={}	\displaystyle \tfrac{5}{2}
\displaystyle -\ \	\displaystyle \bigl(x^2	\displaystyle {}-{}	\displaystyle y^2	\displaystyle {}={}	\displaystyle 2\rlap{\bigr)}
			\displaystyle 2y^2	\displaystyle {}={}	\displaystyle \tfrac{1}{2}

we get \displaystyle y=\pm\tfrac{1}{2}. This gives potentially four solutions to the equation system,

\displaystyle \left\{\begin{align} x &= \tfrac{3}{2}\\[5pt] y &= \tfrac{1}{2} \end{align}\right. \qquad \left\{\begin{align} x &= \tfrac{3}{2}\\[5pt] y &= -\tfrac{1}{2} \end{align}\right. \qquad \left\{\begin{align} x &= -\tfrac{3}{2}\\[5pt] y &= \tfrac{1}{2} \end{align}\right. \qquad \left\{\begin{align} x &= -\tfrac{3}{2}\\[5pt] y &= -\tfrac{1}{2} \end{align}\right.

but only two of these satisfy the second equation \displaystyle 2xy=-\tfrac{3}{2}, namely

\displaystyle \left\{\begin{align} x &= \tfrac{3}{2}\\[5pt] y &= -\tfrac{1}{2} \end{align}\right. \qquad\text{and}\qquad \left\{\begin{align} x &= -\tfrac{3}{2}\\[5pt] y &= \tfrac{1}{2} \end{align}\right.

Hence, we have that the solutions are

\displaystyle w=\frac{3-i}{2}\quad and \displaystyle \quad w=\frac{-3+i}{2}\,,

or, expressed in \displaystyle z,

\displaystyle z=2+i\quad and \displaystyle \quad z=-1+2i\,\textrm{.}

Because the calculation has been rather long, there is the risk that we have calculated incorrectly somewhere and we therefore check that the solutions satisfy the equation in the exercise,

\displaystyle \begin{align} z={}\rlap{2+i:}\phantom{-1+2i:}{}\quad z^2-(1+3i)z-4+3i &= (2+i)^2 - (1+3i)(2+i) - 4 + 3i\\[5pt] &= 4+4i+i^2-(2+i+6i+3i^2)-4+3i\\[5pt] &= 4+4i-1-2-7i+3-4+3i\\[5pt] &= 0\,,\\[10pt] z=-1+2i:\quad z^2-(1+3i)z-4+3i &= (-1+2i)^2-(1+3i)(-1+2i)-4+3i\\[5pt] &= (-1)^2-4i+4i^2-(-1+2i-3i+6i^2)-4+3i\\[5pt] &= 1-4i-4+1+i+6-4+3i\\[5pt] &= 0\,\textrm{.} \end{align}