# Does e^x = 1

1. Sep 26, 2011

### estro

This is not a homework question.

I know that the equation $e^x=1$ has infinitly many solutions in form of $2i\pi n$ but I don't understand how and why.

Will appreciate any help, thanks!

2. Sep 26, 2011

### MaxPlank

Re: e^x=1

You should know that a complex number can be represented as a vector in the gauss plane, where te x-axis is the Real axes and the y-axis is the Imaginary axis. Take for example the complex number z=a+ib (where a and b are real), it can be represented by the vector (a,b). You can use also a polar coordinate system where r is the modulus of the vector (a,b) and $\phi$ is the so-called anomaly, meaning the angle between the vector and the positive side of the x-axis. So you have:

$z=r\left(\cos(\phi)+i\sin(\phi)\right)$

If you consider the taylor expansion of the function sine, cosine and , $e^{x}$you can observe that:

$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$

$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$

and so:

$e^{ix}=\cos(x)+i\sin(x)$

So you can write a complex number with Euler's notation:

$z=re^{i\phi}$

Since in your equation z=1, i.e. the vector (1,0), in Euler's notation you have r=1 and $\phi$ can be 0, $2\pi i$, $4\pi i$ and so on, since they represent the same angle (i.e sine and cosine are periodic and the period is $2\pi$ and so also the exponential is periodic, and the period is $2\pi i$

Last edited by a moderator: Sep 26, 2011
3. Sep 26, 2011

### estro

Re: e^x=1

Thank you for the detailed explanation!

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