Find the complex square roots of 5-12i by solving (x+iy)^2=5-12i for x,y real.

Thanks!

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- Apr 6th 2011, 04:15 AMRine198Complex square roots
Find the complex square roots of 5-12i by solving (x+iy)^2=5-12i for x,y real.

Thanks! - Apr 6th 2011, 04:19 AMFernandoRevilla
- Apr 6th 2011, 04:25 AMRine198
so i've identified that

5=x^2-y^2 (real part)

-12=2xy (imaginary part)

and i tried to do simultaneous equation with it, but i still don't get how i can get the answer? - Apr 6th 2011, 04:28 AMFernandoRevilla
- Apr 6th 2011, 04:55 AMamul28
- Apr 6th 2011, 04:57 AMRine198
but how do i factorise x^4-36-5x^2=0, which is what i got after substituting y=-6/x

- Apr 6th 2011, 05:11 AMe^(i*pi)
Let $\displaystyle u = x^2$ such that $\displaystyle x^4-5x^2-36 = u^2-5u-36 = 0$

This one does factorise - what two number multiply to -36 and sum to -5? - Apr 6th 2011, 05:24 AMRine198
So i've found that x=+/-3 and y=+/-2, but i still don't get how to find the complex square roots from this?

- Apr 6th 2011, 05:34 AMFernandoRevilla
The square roots are $\displaystyle 3-2i,-3+2i$ .

Equivalently : $\displaystyle \pm(3-2i)$ .