Is e^{\pi i} equal to -1 or -e^{-2k\pi^{2}}?

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1. Compute all the values of e^ {\pi i}, indicating clearly whether there is just one or many of them.

Trivially, exp(pi * i) = -1. However, we can also consider e to be the complex number z, and pi * i to be the complex number alpha. Then we get:

e^{\pi i} = z^{\alpha} = e^{\alpha log(z)}<br /> = e^{\alpha (Log |z| + i arg(z))}<br /> = e^{\pi i (Log |e| + i arg(e))}<br /> = e^{\pi i (1 + i2k\pi)}<br /> = e^{\pi i}e^{-2\pi^{2}k}<br /> = - e^{-2\pi^{2}k}<br />
where k is an integer.

So what exactly is going on here? does exp(pi*i) = -1 or -exp(-2kpi^2)??

P.S. I hope all this tex doesn't mess up :(
 
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something is wrong with LaTeX... it isn't displaying my tex right...
 
Fixed your LaTeX.
endeavor said:
1. Compute all the values of e^ {\pi i}, indicating clearly whether there is just one or many of them.

Trivially, exp(pi * i) = -1. However, we can also consider e to be the complex number z, and pi * i to be the complex number alpha. Then we get:

e^{\pi i} = z^{\alpha} = e^{\alpha log(z)}<br /> = e^{\alpha (Log |z| + i arg(z))}<br /> = e^{\pi i (Log |e| + i arg(e))}<br /> = e^{\pi i (1 + i2k\pi)}<br /> = e^{\pi i}e^{-2\pi^{2}k}<br /> = - e^{-2\pi^{2}k}<br />
where k is an integer.

So what exactly is going on here? does exp(pi*i) = -1 or -exp(-2kpi^2)??

P.S. I hope all this tex doesn't mess up :(
 
Is the complex exponential function invertible? (What is required for a function to have an inverse?)
 
NeoDevin said:
Is the complex exponential function invertible? (What is required for a function to have an inverse?)

The function must be 1-1, right?
 
endeavor said:
The function must be 1-1, right?

Correct. Does the complex exponential satisfy this?
 
NeoDevin said:
Correct. Does the complex exponential satisfy this?

Sorry, I misread your first question. So, no, the complex exponential is not an invertible function. Where does my initial post break down then?
 
endeavor said:
Sorry, I misread your first question. So, no, the complex exponential is not an invertible function. Where does my initial post break down then?

When you tried to invert it.
 

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