Originally Posted by

**Henryt999** Trying to solve this, not allowed to use polar form or Eulers.

$\displaystyle z^2-4z+4+2i=0$

Here is my work: $\displaystyle (z-2)^2 = -2i $

then I have to take the root of $\displaystyle -2i$ and that didnīt work very well.

Why? What's the problem? You'll have to do the same if you use the well-known formula for the roots of a quadratic equation: you need the square root of $\displaystyle -2i$ , so you can put:

$\displaystyle (x+yi)^2=-2i\Longleftrightarrow x^2-y^2+2xyi=-2i\Longrightarrow x^2-y^2=0$ , $\displaystyle 2xy=-2\Longrightarrow xy=-1\Longrightarrow y=-\frac{1}{x}$ , so $\displaystyle x^2=y2\Longleftrightarrow x^2=\frac{1}{x^2}\Longrightarrow x^4=1\Longrightarrow x=1\,,\,y=-1\,,\,\,x=-1\,,\,y=1$ are the solutions.

Tonio

I tried this too:

rewritting $\displaystyle z^2 = x^2-y^2+2xyi$

and $\displaystyle -4z = -4x-4yi$

but then it is a $\displaystyle x^4$ equationsystem with two unknows and one x^3 term etc so that didnīt work either.

and here I am, any ideas?