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2bornot2b
Group Title
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.
 2 years ago
 2 years ago
2bornot2b Group Title
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.
 2 years ago
 2 years ago

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2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
@shivam_bhalla
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
\[\text{ Do you recall } e^{ i \theta}= \cos(\theta) +i \sin(\theta)\]
 2 years ago

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

myininaya Group TitleBest ResponseYou've already chosen the best response.0
And of course n is an integer
 2 years ago

myininaya Group TitleBest 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
 2 years ago

2bornot2b Group TitleBest 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.
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
What do I do next @myininaya ?
 2 years ago

KingGeorge Group TitleBest 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.
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
@KingGeorge, is it just substitution? Perhaps you didn't notice I have replaced the ( )
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Herp derp.
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
My ideas aren't panning out :(
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
What do you mean by complex? \(w=x+iy\)?
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
@Ishaan94, Yes.
 2 years ago

Ishaan94 Group TitleBest 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?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
why \( (e^w)^z \not = e^{wz} \) ???
 2 years ago

2bornot2b Group TitleBest 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
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
Certainly you can use euler's identity
 2 years ago

Ishaan94 Group TitleBest 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. \(e^{i\theta} = \cos \theta +i\sin \theta\). \[\sin 30 = \sin150,\,\text{but}\, 150\neq30.\]\[\sin \theta=\sin (2n\pi + \theta)\]
 2 years ago

Ishaan94 Group TitleBest 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.
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
But what does it have to do with this complex identity ?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
hmm, maybe you have to go back to the definition in terms of complex log
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
@estudier Do you mean the definition of \(a^b\) where both a and b are complex?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Problem seems to be you cannot equate a multi value f with a single value f....
 2 years ago

Ishaan94 Group TitleBest 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.
 2 years ago

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

2bornot2b Group TitleBest 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 \(e^{Loga~b}=(e^{Loga})^b\)
 2 years ago

estudier Group TitleBest 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)
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
I really don't like thi stuff.........:(
 2 years ago
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