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Prove that \[\huge{(e^w)^z=e^{[(w+2ni\pi)(z)]}}\] but \[\huge (e^w)^z\ne e^{(wz)}\] where w and z are complex.

Mathematics
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\[\text{ Do you recall } e^{ i \theta}= \cos(\theta) +i \sin(\theta)\]
\[e^{i \theta}=\cos(\theta)+i \sin(\theta)=\cos(\theta+2 n \pi)+ i \sin( \theta + 2 n \pi))=e^{i(\theta+2 n \pi)}\]

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Other answers:

And of course n is an integer
Try to use that in this proof I would write w as a+bi by the way until the end when you write a+bi back as w
So you want me to start with \(\large (e^{a+ib})^z\) Sorry for the late reply @myininaya I was away.
What do I do next @myininaya ?
The first thing you need to show is not too bad. Since \(w\) is complex, \(w=w+2ni\pi \). Hence\[\Large (e^w)^z=e^{(w+2ni\pi)~z}\]by using a simple substitution.
@KingGeorge, is it just substitution? Perhaps you didn't notice I have replaced the ( )
Herp derp.
My ideas aren't panning out :(
What do you mean by complex? \(w=x+iy\)?
@Ishaan94, Yes.
Does the fact that w and z are complex numbers restrict the use of euler's identity?
why \( (e^w)^z \not = e^{wz} \) ???
I don't think so @Ishaan94 @experimentX see bullet three after visiting the following link http://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities
Certainly you can use euler's identity
I know that \(e^{x} =e^{y}\) doesn't imply that x equals y, but I still don't understand it. Hmm okay maybe I know why. \(e^{i\theta} = \cos \theta +i\sin \theta\). \[\sin 30 = \sin150,\,\text{but}\, 150\neq30.\]\[\sin \theta=\sin (2n\pi + \theta)\]
You can only have one unique value of theta in a period by adding 2npi you change the period. Or, you complete the circle and reach the point where you started from.
But what does it have to do with this complex identity ?
hmm, maybe you have to go back to the definition in terms of complex log
@estudier Do you mean the definition of \(a^b\) where both a and b are complex?
Problem seems to be you cannot equate a multi value f with a single value f....
A lot. This is how you represent a complex number in complex plane. |dw:1336309768854:dw|You can only get the same theta value if you complete circle.
z^az^b = e^(alnz)e^(blnz) = e^a(lnz +2pi i n) etc
@estudier Let me tell you the problem I am facing. I was going through the definition of \(a^b\) which is defined as \(e^{b Loga}\). Now I already know that \(a^b=(e^{Loga})^b\) Does that mean \(e^{Loga~b}=(e^{Loga})^b\)
Then z^(a+b) = e^(a+b) ln z = e^(a+b)( ln z + 2 pi i k) = e^(a+b)ln ze^2pi i k(a+b) -> z^(a+b) is in z^az^b (might be same might not)
I really don't like thi stuff.........:-(

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