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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

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myininayaBest ResponseYou've already chosen the best response.0
\[\text{ Do you recall } e^{ i \theta}= \cos(\theta) +i \sin(\theta)\]
 one year ago

myininayaBest ResponseYou've already chosen the best response.0
\[e^{i \theta}=\cos(\theta)+i \sin(\theta)=\cos(\theta+2 n \pi)+ i \sin( \theta + 2 n \pi))=e^{i(\theta+2 n \pi)}\]
 one year ago

myininayaBest ResponseYou've already chosen the best response.0
And of course n is an integer
 one year ago

myininayaBest ResponseYou've already chosen the best response.0
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
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
So you want me to start with \(\large (e^{a+ib})^z\) Sorry for the late reply @myininaya I was away.
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
What do I do next @myininaya ?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
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.
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
@KingGeorge, is it just substitution? Perhaps you didn't notice I have replaced the ( )
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
My ideas aren't panning out :(
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
What do you mean by complex? \(w=x+iy\)?
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
Does the fact that w and z are complex numbers restrict the use of euler's identity?
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
why \( (e^w)^z \not = e^{wz} \) ???
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
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
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
Certainly you can use euler's identity
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
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. \(e^{i\theta} = \cos \theta +i\sin \theta\). \[\sin 30 = \sin150,\,\text{but}\, 150\neq30.\]\[\sin \theta=\sin (2n\pi + \theta)\]
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
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.
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
But what does it have to do with this complex identity ?
 one year ago

estudierBest ResponseYou've already chosen the best response.0
hmm, maybe you have to go back to the definition in terms of complex log
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
@estudier Do you mean the definition of \(a^b\) where both a and b are complex?
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Problem seems to be you cannot equate a multi value f with a single value f....
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
A lot. This is how you represent a complex number in complex plane. dw:1336309768854:dwYou can only get the same theta value if you complete circle.
 one year ago

estudierBest ResponseYou've already chosen the best response.0
z^az^b = e^(alnz)e^(blnz) = e^a(lnz +2pi i n) etc
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
@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 \(e^{Loga~b}=(e^{Loga})^b\)
 one year ago

estudierBest ResponseYou've already chosen the best response.0
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)
 one year ago

estudierBest ResponseYou've already chosen the best response.0
I really don't like thi stuff.........:(
 one year ago
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