Confusion about contour integration of constant function: intuition vs. Residue Theorem

Question

Let's say we have the holomorphic function $$f(z) = 1.$$ Because $f(z)$ has no poles, according the Residue Theorem we have $$\oint_\gamma f(z)\,dz = 0$$ for any closed counterclockwise path $\gamma$.

But let's say that $\gamma$ is a circle around the origin of radius $r$. Then shouldn't we have $$\oint_\gamma f(z)\,dz = 2 \pi r$$ because $$\oint_\gamma f(z)\,dz = \oint_\gamma dz = \text{arclength}\,\gamma$$ ?

I'm pretty sure the result using the Residue Theorem is correct, so then my reasoning must be incorrect for the second way of looking at it.

Where is my reasoning incorrect?

Answer 1

No, because $dz$ does not represent arclength - rather, $|dz|$ does. So the correct statement would be

$$\oint_{\gamma} dz = 0, \quad\quad \oint_{\gamma} |dz| = 2\pi r$$

Remember, you can always go back to the Riemann sum; when defining the integral $dz$, you sum things that look like $\Delta z$. If you move in a circular path, you don't travel anywhere - hence, the sum of $\Delta z$ is zero.

Answer 2

If $f$ is the constant fuction $1$, The integral $$\int_\gamma f(z)\,dz$$ does not give you the length of the arc $\gamma$. That would be $$\int_\gamma f(z)\,d|z|,$$ where $d|z|$ is integration with respect to arc-length.