<s>[INST] Solve a complex equation Solve the following equation $$(4-3i)z^2-25z+31-17i= 0 $$
Dividing by 4-3i gives me $$z^2  \frac{-100z-75zi + 124 + 93i -68i -51i^2}{25}$$ 
which goes to $$z^2 -4z-3zi + 7+i$$
then i collect the terms so $$z - \left(\frac{(4-3i)}{2}\right)^2 = -7 -i + \left(\frac{4-3i}{2}\right)^2$$
and after that i can't get the expansion to work.
Can you help me out? 
 [/INST] Starting from what you ought to have:
  $(z−(2+\frac{3}{2}i))^2 = -7-i + (2+\frac{3}{2}i)^2$.
we get:
  $(z−(2+\frac{3}{2}i))^2 = -7-i + (4+6i-\frac{9}{4}) = \frac{-21+20i}{4} = (\frac{2+5i}{2})^2$.
The last step I obtained by guessing. If you want a systematic way to find the square root of a complex number if it is a perfect square (the square of a rational complex number) you can use the following technique.
Given $(a+bi)^2 = c+di$ where $a,b,c,d \in \mathbb{R}$:
  $a^2-b^2 = c$ and $2ab = d$ [by comparing real and imaginary parts].
  Thus $4 a^2 - 4 a^2 b^2 = 4 a^2 c^2$ and $4 a^2 b^2 = d^2$ [multiply the first by $2a^2$ and square the second].
  Thus $4 a^4 - 4c a^2 - d^2 = 0$ [Add them together to get a quadratic in $a$].
  Thus $(2a^2-c)^2 = c^2+d^2$ [Complete the square].
  Thus $2a^2-c = \sqrt{c^2+d^2}$ and hence $a = \pm \sqrt{\frac{\sqrt{c^2+d^2}+c}{2}}$.
  Also $2b^2 = 2a^2 - 2c = \sqrt{c^2+d^2}-c$ and hence $b = \pm \sqrt{\frac{\sqrt{c^2+d^2}-c}{2}}$.
  [Note that the choices of sign are dependent, so it may be better to use $b = \frac{d}{2a}$ instead.]
In the above case we get:
  $\sqrt{-21+20i} = \pm \left( \sqrt{\frac{\sqrt{21^2+20^2}-21}{2}} + \sqrt{\frac{\sqrt{21^2+20^2}+21}{2}} i \right) = \pm (2+5i)$.
  [Here I have chosen the signs correctly.]
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