# Math Help - Imaginary, quadratic

Trying to solve this, not allowed to use polar form or Eulers.
$z^2-4z+4+2i=0$

Here is my work: $(z-2)^2 = -2i$
then I have to take the root of $-2i$ and that didn´t work very well.

I tried this too:
rewritting $z^2 = x^2-y^2+2xyi$
and $-4z = -4x-4yi$
but then it is a $x^4$ equationsystem with two unknows and one x^3 term etc so that didn´t work either.
and here I am, any ideas?

2. Originally Posted by Henryt999
Trying to solve this, not allowed to use polar form or Eulers.
$z^2-4z+4+2i=0$

Here is my work: $(z-2)^2 = -2i$
then I have to take the root of $-2i$ and that didn´t 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 $-2i$ , so you can put:

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

Tonio

I tried this too:
rewritting $z^2 = x^2-y^2+2xyi$
and $-4z = -4x-4yi$
but then it is a $x^4$ equationsystem with two unknows and one x^3 term etc so that didn´t work either.
and here I am, any ideas?
.

3. First of all thank you for your help.
But are you sure this is correct!
Entering $x = 1; y=(-1)$ into the $z^2-4z+4+2i = 0$ doesnt = 0
or maybe I made some mistake?:

4. Originally Posted by Henryt999
First of all thank you for your help.
But are you sure this is correct!
Entering $x = 1; y=(-1)$ into the $z^2-4z+4+2i = 0$ doesnt = 0
or maybe I made some mistake?:

Who said $x=1\,,\,y=-1\Longleftrightarrow z=1-i$ is a solution to your question?!? . It is a solution to $z^2=-2i$ !! Now take this, input it in the right place and THEN find the correcto solutions! Just re-read your own intent of solution ...

Tonio