1. ## complex numbers

Hi is my reasoning correct?

I want z^2=conj(Z) in the form z=a+bi.

What I think is z=sqrt[conj(Z)].

Then

Z= sqrt[a-bi].

Z=sqrt(a) – sqrt(b)i.

Thanks………………

2. I would multiply out the LHS thus:

(a+bi)(a+bi)= a^2 + 2abi - b^2, and equate real and imaginary components to the RHS: a-bi. Two equations and two unknowns.

3. ## response

So if

(a+bi)(a+bi)= a^2 + 2abi - b^2

Then

Z= a^2 + 2abi or a(a +2bi)

or

Z= -(b)^2 + 2abi or -b(b -2ai)

4. Re your first post: in general $\sqrt{a+ b}\ne \sqrt{a}+ \sqrt{b}$. Also are "z" and "Z" the same? It is bad practice to use capital and non-capital letters interchangeably.

I have no idea where you got "Z= a^2+ 2abi". You seem to have just dropped the "-b^2" from the previous line. And then just dropped "a^2" to get that last line. You can't do that!

If z= a+ bi, then z^2= a^2- b^2+ 2abi. conj(z)= a- bi so your equation, z^2= conj(z), is a^2- b^2+ 2abi= a- bi. Equating real and imaginary parts gives a^2- b^2= a and 2ab= -b.

I get four complex number solutions.