Thread: Complex square roots

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

Thanks!

Originally Posted by Rine198

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

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

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?

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 , substitute in the first equation and you'll obtain a biquadratic equation on .

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.

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

Let such that

This one does factorise - what two number multiply to -36 and sum to -5?

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?

The square roots are .

Equivalently : .