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I am trying to prove $\text{Ln}\,z_1^{z_2}=z_2\,\text{Ln}\,z_1$. If I know that $(e^{z_1})^{z_2}=e^{z_1z_2}$, then it would be:

Let $z_1=e^w$, then $z_1^{z_2}=(e^w)^{z_2}=e^{wz_2}$. Take Ln on both side, hence Ln$\,z_1^{z_2}=wz_2=z_2\,\text{Ln}\,z_1$.

But now I have another problem: how to prove that $(e^{z_1})^{z_2}=e^{z_1z_2}$ ?

Thank you.

Gary
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  • How to prove these statements depends on the definitions you start with, would you supply those? – Christoph Jan 25 '14 at 15:41
  • Along with @ChristophPegel's comment, you need to start off with an expression on the left hand side or right hand side (usually, a mathematician starts proving the right hand side from the left hand side). For instance, I am given $\mathrm{Ln}z_1^{z_2}$. I need to show that it's equivalent to $z_2 \mathrm{Ln}z_1$. – NasuSama Jan 25 '14 at 15:44
  • $e^z$ is defined to be $\sum_{k=0}^{\infty}\frac{z^k}{k!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\dots$. But direct substitution seems very complicated. – Gary Jan 25 '14 at 15:45
  • How do you define $z^w$ for $z,w\in\mathbb C$? – Christoph Jan 25 '14 at 15:48
  • OMG. Thanks for you help. Now I understand: $z^w$ is defined by $e^{w\text{Ln},z}$, so Ln$,z^w=w\text{Ln},z$. Hence $(e^z)^w=e^{zw}$. Thanks. – Gary Jan 25 '14 at 15:58

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It's not true. (Assuming you mean the principal branch of the complex logarithm, but it's not true for any other specific branch either.)

Take $z_1=-1+i$ and $z_2=2$. Then $z_1^{z_2} = -2i$ so $\operatorname{Log} z_1^{z_2} = \ln 2 - i\pi/2$.

On the other hand $2\operatorname{Log} (-1+i) = 2( \ln\sqrt2 + 3i\pi/4) = \ln 2 + 3i\pi/2$.

mrf
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  • But Ln$,-2i$ is multivalued, so it should be Ln$,z_1^{z_2}=\ln2+{i(2k+3/2)\pi}$, where $k$ is integer. As I stated in the above comment, it should be true, because as $z^w$ is defined by $e^{w\text{Ln},z}$, so Ln$,z^w=\text{Ln},e^{w\text{Ln},z}=w\text{Ln},z$. Hence $(e^z)^w=e^{zw}$. – Gary Jan 25 '14 at 16:12
  • If you meant the multivalued interpretation, you should really have mentioned that. Especially after pepople asked about your definition. The most common convention is that capitalized $\operatorname{Log}$ refers to the prinipal branch. – mrf Jan 25 '14 at 16:17
  • The book I am using (Mathematical Methods for Physics and Engineering (Riley)) really uses Ln to represent the multivalued interpretation, but I rarely see its use in other textbooks. Anyway sorry for the confusion and thanks for your help. – Gary Jan 25 '14 at 16:22