Complex numbers and completing the square

[HMPARTICLE]

Homework Statement

let z' = (a,b), find z in C such that z^2 = z'

Homework Equations

The Attempt at a Solution

let z = (x,y) then z^2 = (x^2-y^2, 2xy)
since z^2 = z', we have,

(x^2-y^2, 2xy) = (a,b)

comparing real and imaginary components we have;

x^2-y^2 = a,
2xy = b.

Now, this is where i am stuck.
i know i have to find z in terms of a and b.
here is an attempt at what to do next;

subtracting the second equation from the first,

x^2 - 2xy - y^2 = a - b

completing the square,

(x - y)^2 - 2y^2 = a - b.

...

Not asking for the answer or anything, just a push in the right direction.

 

[jedishrfu]

You have z' as a+bi and you know then that z is a-bi so shouldn't you just solve for a in terms of b or vice versa?

 

[Ray Vickson]

Homework Statement

let z' = (a,b), find z in C such that z^2 = z'

Homework Equations

The Attempt at a Solution

let z = (x,y) then z^2 = (x^2-y^2, 2xy)
since z^2 = z', we have,

(x^2-y^2, 2xy) = (a,b)

comparing real and imaginary components we have;

x^2-y^2 = a,
2xy = b.

Now, this is where i am stuck.
i know i have to find z in terms of a and b.
here is an attempt at what to do next;

subtracting the second equation from the first,

x^2 - 2xy - y^2 = a - b

completing the square,

(x - y)^2 - 2y^2 = a - b.

...

Not asking for the answer or anything, just a push in the right direction.

Write $z' = a + ib$ in polar form. Then, finding the roots of $z^2 = z'$ is straightforward.

 

[Joffan]

It might be worth calculating |z| and |z²|. Also investigating what (kz)² looks like and thinking about the implications for z.

 

[HMPARTICLE]

Thanks guys! My textbook hadn't yet defined polar form so I was reluctant to use it. I solved in the way I was progressing via substitution.

 

[HallsofIvy]

I don't think it helps to convert to polar form. What you already have looks great to me except that you have written z as x+ yi and z' as a+ bi. So exactly what does z' mean here? You are treating it as just "some other" complex number but I would have interpreted it as the complex conjugate of z. If that is correct then the problem is much easier than what you are doing.

 

[HMPARTICLE]

no, some other complex number (not complex conjugate), i should have denoted it as $z_0$ or something similar!