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.

Question

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.

2bornot2b · Answer

@shivam_bhalla

myininaya · Answer

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

myininaya · Answer

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

myininaya · Answer

And of course n is an integer

myininaya · Answer

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 · Answer

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

2bornot2b · Answer

What do I do next @myininaya ?

kinggeorge · Answer

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 · Answer

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

kinggeorge · Answer

Herp derp.

kinggeorge · Answer

My ideas aren't panning out :(

anonymous · Answer

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

2bornot2b · Answer

@Ishaan94, Yes.

anonymous · Answer

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

experimentx · Answer

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

2bornot2b · Answer

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 · Answer

Certainly you can use euler's identity

anonymous · Answer

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 · Answer

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 · Answer

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

anonymous · Answer

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

2bornot2b · Answer

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

anonymous · Answer

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

anonymous · Answer

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.