Exploring e^{\pi i}: Is It -1 or -e^{-2k\pi^{2}}?

[endeavor]

[tex][/tex]

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:

$$e^{\pi i} = z^{\alpha} = e^{\alpha log(z)} = e^{\alpha (Log |z| + i arg(z))} = e^{\pi i (Log |e| + i arg(e))} = e^{\pi i (1 + i2k\pi)} = e^{\pi i}e^{-2\pi^{2}k} = - e^{-2\pi^{2}k}$$
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 :(

 

[endeavor]

something is wrong with LaTeX... it isn't displaying my tex right...

 

[Mark44]

Fixed your LaTeX.

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:

$$e^{\pi i} = z^{\alpha} = e^{\alpha log(z)} = e^{\alpha (Log |z| + i arg(z))} = e^{\pi i (Log |e| + i arg(e))} = e^{\pi i (1 + i2k\pi)} = e^{\pi i}e^{-2\pi^{2}k} = - e^{-2\pi^{2}k}$$
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 :(

 

[NeoDevin]

Is the complex exponential function invertible? (What is required for a function to have an inverse?)

 

[endeavor]

Is the complex exponential function invertible? (What is required for a function to have an inverse?)

The function must be 1-1, right?

 

[NeoDevin]

The function must be 1-1, right?

Correct. Does the complex exponential satisfy this?

 

[endeavor]

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?

 

[NeoDevin]

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.