## 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

1. 2bornot2b Group Title

@shivam_bhalla

2. myininaya Group Title

$\text{ Do you recall } e^{ i \theta}= \cos(\theta) +i \sin(\theta)$

3. myininaya Group Title

$e^{i \theta}=\cos(\theta)+i \sin(\theta)=\cos(\theta+2 n \pi)+ i \sin( \theta + 2 n \pi))=e^{i(\theta+2 n \pi)}$

4. myininaya Group Title

And of course n is an integer

5. myininaya Group Title

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

6. 2bornot2b Group Title

So you want me to start with $$\large (e^{a+ib})^z$$ Sorry for the late reply @myininaya I was away.

7. 2bornot2b Group Title

What do I do next @myininaya ?

8. KingGeorge Group Title

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.

9. 2bornot2b Group Title

@KingGeorge, is it just substitution? Perhaps you didn't notice I have replaced the ( )

10. KingGeorge Group Title

Herp derp.

11. KingGeorge Group Title

My ideas aren't panning out :(

12. Ishaan94 Group Title

What do you mean by complex? $$w=x+iy$$?

13. 2bornot2b Group Title

@Ishaan94, Yes.

14. Ishaan94 Group Title

Does the fact that w and z are complex numbers restrict the use of euler's identity?

15. experimentX Group Title

why $$(e^w)^z \not = e^{wz}$$ ???

16. 2bornot2b Group Title

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

17. 2bornot2b Group Title

Certainly you can use euler's identity

18. Ishaan94 Group Title

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)$

19. Ishaan94 Group Title

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.

20. 2bornot2b Group Title

But what does it have to do with this complex identity ?

21. estudier Group Title

hmm, maybe you have to go back to the definition in terms of complex log

22. 2bornot2b Group Title

@estudier Do you mean the definition of $$a^b$$ where both a and b are complex?

23. estudier Group Title

Problem seems to be you cannot equate a multi value f with a single value f....

24. Ishaan94 Group Title

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.

25. estudier Group Title

z^az^b = e^(alnz)e^(blnz) = e^a(lnz +2pi i n) etc

26. 2bornot2b Group Title

@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 $$e^{Loga~b}=(e^{Loga})^b$$

27. estudier Group Title

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)

28. estudier Group Title

I really don't like thi stuff.........:-(