1. ## Complex square roots

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

Thanks!

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

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

Now, identify real and imaginary parts and solve the system.

3. 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?

4. 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 $\displaystyle y=-6/x$, substitute in the first equation and you'll obtain a biquadratic equation on $\displaystyle x$ .

5. 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.

6. but how do i factorise x^4-36-5x^2=0, which is what i got after substituting y=-6/x

7. 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?

8. 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?

9. The square roots are $\displaystyle 3-2i,-3+2i$ .

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