Existence of (complex) limit z->0 (z^a)

1. Sep 22, 2009

Knissp

1. The problem statement, all variables and given/known data
Justify for which complex values of a does the principal value of $$z^a$$ have a limit as z tends to 0?

2. Relevant equations

$$z^a = e^{a log(z)}$$

$$log(z) = log|z| + (i) (arg(z))$$

3. The attempt at a solution

$$Lim_{z \rightarrow 0} z^a = Lim_{z \rightarrow 0} e^{(a) (log(z))}$$

$$=Lim_{|z| \rightarrow 0} e^{(a) (log|z|) + (i) (a) (arg(z))}$$

Let $$a = u + i v$$.

$$=Lim_{|z| \rightarrow 0} e^{(u+iv) (log|z| + (i) (u+iv) (arg(z)))}$$

$$=Lim_{|z| \rightarrow 0} e^{(u) (log|z|) + (i) (v) (log|z|) + (i) (u) (arg(z)) - (v) (arg(z))}$$

$$=Lim_{|z| \rightarrow 0} e^{(u) (log|z|)} e^{(i) (v) (log|z|)} e^{(i) (u) (arg(z))} e^{-v (arg(z))}$$

$$=Lim_{|z| \rightarrow 0} |z|^u e^{(i) (v) (log|z|)} e^{((i) (u) - (v)) (arg(z))}$$

I just noticed a big mistake here, so I'm erasing this part. Any ideas?

Last edited: Sep 22, 2009
2. Sep 22, 2009

tiny-tim

Hi Knissp!

Isn't it easier just to go polar, and put z = re ?