# Complex square roots

• Apr 6th 2011, 04:15 AM
Rine198
Complex 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 AM
FernandoRevilla
Quote:

Originally Posted by Rine198
Find the complex square roots of 5-12i by solving (x+iy)^2=5-12i for x,y real.

$(x+iy)^2=x^2-y^2+2xyi=5-12i$

Now, identify real and imaginary parts and solve the system.
• Apr 6th 2011, 04:25 AM
Rine198
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 AM
FernandoRevilla
Quote:

Originally Posted by Rine198
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?

Right, now $y=-6/x$, substitute in the first equation and you'll obtain a biquadratic equation on $x$ .
• Apr 6th 2011, 04:55 AM
amul28
Quote:

Originally Posted by Rine198
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?

Show what you have done.
• Apr 6th 2011, 04:57 AM
Rine198
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 AM
e^(i*pi)
Let $u = x^2$ such that $x^4-5x^2-36 = u^2-5u-36 = 0$

This one does factorise - what two number multiply to -36 and sum to -5?
• Apr 6th 2011, 05:24 AM
Rine198
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 AM
FernandoRevilla
The square roots are $3-2i,-3+2i$ .

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