2bornot2b
  • 2bornot2b
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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2bornot2b
  • 2bornot2b
@shivam_bhalla
myininaya
  • myininaya
\[\text{ Do you recall } e^{ i \theta}= \cos(\theta) +i \sin(\theta)\]
myininaya
  • myininaya
\[e^{i \theta}=\cos(\theta)+i \sin(\theta)=\cos(\theta+2 n \pi)+ i \sin( \theta + 2 n \pi))=e^{i(\theta+2 n \pi)}\]

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myininaya
  • myininaya
And of course n is an integer
myininaya
  • myininaya
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
2bornot2b
  • 2bornot2b
So you want me to start with \(\large (e^{a+ib})^z\) Sorry for the late reply @myininaya I was away.
2bornot2b
  • 2bornot2b
What do I do next @myininaya ?
KingGeorge
  • KingGeorge
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.
2bornot2b
  • 2bornot2b
@KingGeorge, is it just substitution? Perhaps you didn't notice I have replaced the ( )
KingGeorge
  • KingGeorge
Herp derp.
KingGeorge
  • KingGeorge
My ideas aren't panning out :(
anonymous
  • anonymous
What do you mean by complex? \(w=x+iy\)?
2bornot2b
  • 2bornot2b
@Ishaan94, Yes.
anonymous
  • anonymous
Does the fact that w and z are complex numbers restrict the use of euler's identity?
experimentX
  • experimentX
why \( (e^w)^z \not = e^{wz} \) ???
2bornot2b
  • 2bornot2b
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
2bornot2b
  • 2bornot2b
Certainly you can use euler's identity
anonymous
  • anonymous
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. \(e^{i\theta} = \cos \theta +i\sin \theta\). \[\sin 30 = \sin150,\,\text{but}\, 150\neq30.\]\[\sin \theta=\sin (2n\pi + \theta)\]
anonymous
  • anonymous
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.
2bornot2b
  • 2bornot2b
But what does it have to do with this complex identity ?
anonymous
  • anonymous
hmm, maybe you have to go back to the definition in terms of complex log
2bornot2b
  • 2bornot2b
@estudier Do you mean the definition of \(a^b\) where both a and b are complex?
anonymous
  • anonymous
Problem seems to be you cannot equate a multi value f with a single value f....
anonymous
  • anonymous
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.
anonymous
  • anonymous
z^az^b = e^(alnz)e^(blnz) = e^a(lnz +2pi i n) etc
2bornot2b
  • 2bornot2b
@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 \(e^{Loga~b}=(e^{Loga})^b\)
anonymous
  • anonymous
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)
anonymous
  • anonymous
I really don't like thi stuff.........:-(

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