## 2bornot2b 3 years 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.

1. 2bornot2b

@shivam_bhalla

2. myininaya

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

3. myininaya

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

4. myininaya

And of course n is an integer

5. 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

6. 2bornot2b

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

7. 2bornot2b

What do I do next @myininaya ?

8. 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.

9. 2bornot2b

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

10. KingGeorge

Herp derp.

11. KingGeorge

My ideas aren't panning out :(

12. Ishaan94

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

13. 2bornot2b

@Ishaan94, Yes.

14. Ishaan94

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

15. experimentX

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

16. 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

17. 2bornot2b

Certainly you can use euler's identity

18. Ishaan94

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

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

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

21. estudier

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

22. 2bornot2b

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

23. estudier

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

24. Ishaan94

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

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

26. 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 $$e^{Loga~b}=(e^{Loga})^b$$

27. estudier

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

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