Problem with square roots of complex number

I have to tackle a question related with complex number and its square roots. My thoughts so far are below it.

(a)Given that (x+iy)^2=-5+12i,x and y belongs to real numbers, show that:

  (i)x^2-y^2=-5
  (ii)xy=6

It is easy for me to solve this first question, and what I only need to do do is extending the left side of the equation and get the results.
But what really makes me confused is the rest of the question:

(b)Hence find the two square roots of -5+12i

How can I find a root of a complex number? Using the viete’s formula?
What I only know is to find z from z^n=rcosθ.

Then it asks:

(C)for any complex number z, show that (z*)^2=(z^2)^*

(D)hence write down the two square roots of -5-12i

I thought C is easy to prove.Maybe I can suppose z=a+bi and z^*=a-bi and plug them into the equation provided.But how can it helps to find the roots??

How can I solve this problem ?? Help

Answer

Let $z=x+iy$. For $(b)$, you need to solve $x^2-y^2=-5$ and $xy=6$. This is not too difficult to solve using Theo Bendit's answer but a nice trick is to remark that

$$x^2+y^2=|z|^2=|z^2|=|-5+12i|=13$$

Call this equation (iii). (i) and (iii) are linear in $x^2$ and $y^2$, so the system they form is easy to solve. We obtain $x^2=4$ and $y^2=9$, so $x=\pm2$ and $y=\pm3$.

This gives for possibilities for $(x,y)$ but only 2 are solutions of the problem since $xy=6$. Therefore, there are two solutions: $(x,y)=(2,3)$ and $(x,y)=(-2,-3)$. In other words, the square roots of $-5+12i$ are $2+3i$ and $-2-3i$.