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For example, when $\alpha = 2$, $\ln(z^{2}) \neq 2\ln(z)$, because argument z is determined up to constant $2 \pi k$. So

$$ \ln(z^{2}) = \ln(z) + \ln(z) = \ln(z_{k_{1}}) + \ln(z_{k_{2}}) \neq 2\ln(z_{k_{3}}). $$

Of course, two of correct answers - $1, -1$. But I know, that there are an infinite number of answers for $\alpha$. Can you help me?

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What is your definition of $z^\alpha$, given complex numbers $z$ and $\alpha\ $? – Christian Blatter Sep 25 '12 at 11:32
up vote 1 down vote accepted

Using this and this,

$Log (a+ib)^{x+iy}$ $=\frac{1}{2}xlog(a^2+b^2)-y(2m\pi+tan^{-1}\frac{b}{a})+i(\frac{1}{2}ylog(a^2+b^2)+x(2m\pi+\tan^{-1}\frac{b}{a}))$

$Log (a+ib)= \frac{1}{2}log(a^2+b^2)+i(2n\pi+\tan^{-1}\frac{b}{a})$

$(x+iy)Log (a+ib)$ $=\frac{1}{2}xlog(a^2+b^2)-y(2n\pi+tan^{-1}\frac{b}{a})+i(x(2n\pi+\tan^{-1}\frac{b}{a})+\frac{1}{2}ylog(a^2+b^2))$

The principal values will be same if $-\pi<\frac{1}{2}y\log(a^2+b^2)+x\tan^{-1}\frac{b}{a}\le \pi$ else $2r\pi$ (where $r$ is any integer) must be added to this argument to adjust its value in $(-\pi, \pi]$.

If $x=0$ or $(b=0$ and $a>0 ⇔ \tan^{-1}\frac{b}{a}=0)$ , the condition becomes $-\pi<\frac{1}{2}y\log(a^2+b^2)\le \pi$

If $y=0$ or $a^2+b^2=1 ⇔ \log(a^2+b^2)=0$, the condition becomes $-\pi<x\tan^{-1}\frac{b}{a}\le \pi$

If $y=0$ or $a^2+b^2=1$ and $x=-1$ the condition becomes $-\pi<-\tan^{-1}\frac{b}{a}\le \pi\implies \pi>\tan^{-1}\frac{b}{a}\ge -\tan^{-1}\frac{b}{a}$ which is true for all $a,b$.

If $y=0,x=2$ the condition becomes $-\pi<2\tan^{-1}\frac{b}{a}\le \pi $

$\implies -\frac{\pi}2<\tan^{-1}\frac{b}{a}\le \frac{\pi}2$ i.e, $a>0$.

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Thank you! But how did you get a condition $$ -\pi < \frac{1}{2}y ln(a^{2} + b^{2}) + x arctg\left(\frac{b}{a}\right) \leqslant \pi ? $$ – John Taylor Sep 26 '12 at 3:56
Please look into… and – lab bhattacharjee Sep 26 '12 at 5:23
Thank you! You helped me very much. – John Taylor Sep 27 '12 at 18:07
There is another solution of this problem, as I think. For simplifying I make for $\alpha$ value $x = Real$. So for $z = re^{i(\varphi + 2 \pi k)}$ I can earn for multivalued $Ln(z^{\alpha})$ $$ Ln(z^{\alpha}) = \alpha\left(i(\varphi + 2 \pi k) + ln(r)\right) + 2 \pi j i = \alpha \left(i(\varphi + 2 \pi k) + ln(r) \right) + 2 \pi j i \qquad (.1), $$ and for $\alpha Ln(z)$ $$ \alpha(Ln(z)) = \alpha (i(\varphi + 2 \pi n) + ln(r)) . $$ So $$ \alpha (Ln(z)) = Ln(z^{\alpha}) \Rightarrow |(.1)| \Rightarrow 2 \pi i ( j + \alpha k ) = \alpha 2 \pi i n. $$ – John Taylor Sep 28 '12 at 16:14
So, $\alpha = \frac{1}{n}, n$ - any whole number. – John Taylor Sep 28 '12 at 16:15

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