An algebraic property of complex numbers

Click For Summary
For any complex number z in ℂ\{0}, the relationship |z^{-1/2}|^2 = |z^{-1}| = |z|^{-1} = 1/|z| holds true. Proving this can be challenging, and using polar form for z may simplify the process. It's important to note that the square root function is not single-valued in the complex plane, as each complex number has two square roots, leading to potential ambiguities. Additionally, specifying that z is non-zero is crucial, as some instructors may penalize for omitting this detail. Overall, clarity on these points is essential for accurate mathematical communication.
AxiomOfChoice
Messages
531
Reaction score
1
I'm guessing that if z\in \mathbb C, then we have

<br /> \left| z^{-1/2} \right|^2 = |z^{-1}| = |z|^{-1} = \frac{1}{|z|}.<br />

Proving this seems to be a real headache though. Is there a quick/easy way to do this?
 
Physics news on Phys.org
write z in polar form?
 
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 z1/2 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 :) .
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K