An algebraic property of complex numbers

AxiomOfChoice

I'm guessing that if [itex]z\in \mathbb C[/itex], then we have

[tex]
\left| z^{-1/2} \right|^2 = |z^{-1}| = |z|^{-1} = \frac{1}{|z|}.
[/tex]

Proving this seems to be a real headache though. Is there a quick/easy way to do this?

qbert

write z in polar form?

Bacle2

Don't mean to nitpick, but remember that it is for z in ℂ\{0} to start with; some profs.

may take away points in an exam if you don't specify this.

But also, remember your square root is not defined everywhere, at least not as a function,

but as a multifunction, since every complex number has two square roots. I mean, the

expression z^1/2 is ambiguous until you choose a branch.

Sorry if you already are taking this into account; I am in nitpicking mode, but I

shouldn't take it out on you :) .

micromass

Several posts discussing square roots have been copied to their own thread: http://www.physicsforums.com/showthread.php?t=626736