Thread: Modulus of the square of a complex number ??!?

1. Modulus of the square of a complex number ??!?

Hi, I've been asked to show the following for a complex number z Attempt:

If I let z = a +bi and do the algebra I end up with

a2 - b2 +2abi / a2 + b2 = 1

And I can't see how this could always be true?

If I follow this through I get ai = b ?

Help would be greatly appreciated this is driving me nuts! thanks in advance :-)

2. Re: Modulus of the square of a complex number ??!?

in general this isn't true which should be fairly obvious as $z^2$ can have an imaginary part whereas $|z^2|$ is always real (and non-negative)

3. Re: Modulus of the square of a complex number ??!?

I suspect that the problem was supposed to be to show that $\displaystyle \frac{|z|^2}{|z^2|}= 1$.

4. Re: Modulus of the square of a complex number ??!?

This has sat for a couple of days now so: writing z= a+ bi, $\displaystyle |z|= \sqrt{a^2+ b^2}$ so $\displaystyle |z|^2= a^2+ b^2$. On the other hand $\displaystyle z^2= (a+ bi)^2= a^2- b^2+ (2ab)i$ so $\displaystyle |z^2= \sqrt{(a^2- b^2)^2+ 4a^2b^2}= \sqrt{a^4- 2a^2b^2+ b^4+4a^2b^2}$$\displaystyle = \sqrt{a^4+ 2a^2b^2+ b^4}= \sqrt{(a^2+ b^2)^2}= a^2+ b^2$ also.

Almost trivial would be to write z in polar form: $\displaystyle z= re^{i\theta}$. Then |z|= r so that $\displaystyle |z|^2= r^2$ and $\displaystyle z^2= r^2e^{2i\theta}$ and $\displaystyle |z^2|= r^2$ also.