 Averi Mitchell

2022-06-25

Why can a Complex Logarithm have infinitely many values?
What does this mean, that "Due to the periodicity of the trigonometric functions, a complex logarithm can have infinitely many values"?
$\mathrm{ln}z=\mathrm{ln}\left(\mathrm{cos}x+i\mathrm{sin}x\right)=?$ Elianna Douglas

Expert

The complex logarithm is the inverse of the complex exponential. By Euler's formula,
${e}^{iy}=\mathrm{cos}y+i\mathrm{sin}y.$
Hence for any integer n,
${e}^{x+\left(y+2\pi n\right)i}={e}^{x}\left(\mathrm{cos}y+i\mathrm{sin}\left(y+2\pi n\right)\right)={e}^{x}\left(\mathrm{cos}y+i\mathrm{sin}y\right)={e}^{x+iy}.$
So the complex exponential takes each of its values infinitely often, from which it follows that the complex logarithm has infinitely many values (for this reason, we much choose a branch cut for the complex logarithm). Feinsn

Expert

The exponential function $f\left(x\right)={e}^{x}$ as a function that takes real numbers to positive real numbers, is a one-to-one and onto function, and therefore has a well-defined inverse function: for each positive y, there is just one number (denoted $\mathrm{log}\left(y\right)$) such that ${e}^{\mathrm{log}\left(y\right)}=y$
But the exponential function $f\left(z\right)={e}^{z}=\sum _{n\ge 0}\frac{{z}^{n}}{n!}$ considered as a function on the complex numbers, is not one-to-one. There is in particular the wonderful identity due to Euler,
${e}^{ix}=\mathrm{cos}\left(x\right)+i\mathrm{sin}\left(x\right),$
which shows for example that there are infinitely many solutions to the equation ${e}^{z}=1$ namely $z=2\pi k$ for $k=\dots ,-3,-2,-1,0,1,2,3\dots$ So the complex exponential function is not one-to-one.
More fundamentally, while we could just randomly choose a solution $w$ to the equation ${e}^{w}=z$ for each z, and call those choices collectively "the logarithm of $z$ there is no way to choose so that the result will give a continuous logarithm function.

Do you have a similar question?