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my textbook [H. A. Priestley - Introduction to Complex Analysis] states about the argument of a complex number raised to a power :

'Only when $\alpha$ is an integer does $[z^{\alpha}]$not produce multiple values: in this case $[z^{\alpha}]$ contains the single point $z^{\alpha}$'

I don't understand this statement. Surely, in all cases we have:

$$z^{\alpha}=e^{\alpha(ln|z|+i\theta)}:\theta \in [arg\ z]=\alpha [principal\ argument\ of\ z + 2k\pi,\ k \in \Bbb Z]$$

so that for example $arg[i^2] = [2 (\frac{\pi}{2}+ + 2k\pi,\ k \in \Bbb Z)]=[\pi+2k\pi, k\in \Bbb Z] \notin [\pi]$

Why is this any less multi-valued than a non-integer power ?

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  • $\begingroup$ Because the fractional part of an integer is $0$. $\endgroup$ – Lucian Jun 21 '15 at 13:57
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Let me fix some notation. Let $\theta_0$ be the principal argument of $z$. Write $z^\alpha = e^{\alpha(\ln|z| + i \theta)}$. The argument of $z^\alpha$ can be found from the factor $e^{i\alpha \theta}$ since $\alpha$ is real. The multiple values would normally enter here, since we have multiple choices for $\theta$, namely $\theta = \theta_0 + 2\pi k$, $k\in \mathbb{Z}$. However, note that $$ e^{i\alpha \theta} = e^{i \alpha \theta_0} e^{2\pi i \alpha k} = e^{i\alpha \theta_0}. $$ This is true precisely because $\alpha$ is an integer, allowing $e^{2\pi i \alpha k }=1$. So, all possible choices of phase lead to the same output for $z^\alpha$. We see then that when $\alpha$ is not an integer, $e^{2\pi i \alpha k }$ may not be equal to $1$, and we can get multiple values depending on the value of $k$.

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    $\begingroup$ now I understand thank you! $\endgroup$ – user3203476 Jun 21 '15 at 5:44

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