An algebraic property of complex numbers

[AxiomOfChoice]

I'm guessing that if $z\in \mathbb C$, then we have

$$\left| z^{-1/2} \right|^2 = |z^{-1}| = |z|^{-1} = \frac{1}{|z|}.$$

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: https://www.physicsforums.com/showthread.php?t=626736

 

[blue_raver22]

Yes, there is a quick and easy way to prove this property. We can use the fact that the absolute value of a complex number z is equal to the square root of the product of z and its complex conjugate, denoted as z*.

So, let's start by writing z in polar form as z = re^(iθ). Then, z* = re^(-iθ). Using this, we can rewrite the expression as:

|z^{-1/2}|^2 = \left| \frac{1}{\sqrt{r} e^{i\frac{\theta}{2}}} \right|^2

= \frac{1}{r} \left| e^{-i\frac{\theta}{2}} \right|^2

= \frac{1}{r} e^{-i\frac{\theta}{2}} e^{i\frac{\theta}{2}}

= \frac{1}{r}

= |z|^{-1}

Therefore, we have shown that |z^{-1/2}|^2 = |z|^{-1}, which also implies that |z^{-1}| = |z|^{-1}. This is a useful property that can be used in various applications involving complex numbers.