# Why is $\log(e^z)\ne z$ in complex analysis?

I always know that the natural logarithm $\log_e$ is the inverse of the exponential function $e$, but to my surprise when reading elementary functions in complex analysis, I discover that that is not always true. $$\log(e^z)=z+2k\pi i$$

My question here is that, why is the above expression true?

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Because the exponential function is $2\pi i$-periodic, which implies that it is not one-to-one. This means that given the value of $e^z$ one cannot uniquely determine what $z$ is. – arjafi Mar 9 '12 at 5:56
$\cos(\theta) = \cos(\theta + 2k\pi).$ – user2468 Mar 9 '12 at 6:09

The better question perhaps, is why is $e^z$ periodic?

Well this has somewhat of a geometric interpretation, and for simplicity lets stick with a complex number of the form $i \theta$. Then:

$$e^{i \theta} = \left( \cos \theta + i \sin \theta \right)$$

and you can think of this as identifying the point $(1,\theta)$ on the unit circle in the complex plane (this is polar coordinates). But then of course if you add $2 \pi$ to this angle, you just get back to the same point on the circle. Thus the complex exponential function cannot be injective.

Thus if you want the logarithm of $w$ to be a complex number $z$ such that $e^z=w$, then this will also carry a periodicity.

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The logarithm is not well-defined, because a function must be one-to-one to have an inverse. Since $e^{z}=e^{z+2\pi i}$, the exponential function is not one-to-one. We sometimes define a complex logarithm function by making a choice, for example we could insist that the imaginary part of $\log(z)$ is in the interval $[0,2\pi)$ for all $z$. But for any such choice, $\log(e^z)$ will differ from $z$ by a multiple of $2\pi i$ for most values of $z$. This is unavoidable, since $e^{z}=e^{z+2\pi i}$ and thus $\log(e^z)=\log(e^{z+2\pi i})$.

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