Converse of Cauchy's Theorem (special case) Let $\gamma_0, \gamma_1$ denote two curves in $\mathbb{C}-\{0\}$ that both start and end at 1. I want to show that if
\begin{equation}
\int_{\gamma_0} \frac{1}{z} \ dz = \int_{\gamma_1} \frac{1}{z} \ dz,
\end{equation}
then $\gamma_0$ and $\gamma_1$ are homotopic. Since the above integrals take the values $2\pi i k$ (depending on how many times the curve circles $0$). The statement makes perfectly sense to me (just by drawing some curves around $0$), but I do now know how to argue it properly. Any hints?
 A: Without loss of generality, you can assume that both $\gamma_0$ and $\gamma_1$ have their values in $S^1 = \{z \in \mathbb{C}, |z|^2 = 1\}$. This can indeed be achieved through an homotopy that projects radially each point of $\mathbb{C} \setminus \{0\}$ to the corresponding point of the sphere.
So $\gamma_0$ and $\gamma_1$ are path from $[0,1]$ to $S^1$ such that $\gamma_i(0) = \gamma_i(1) = 1$ for $i = 0, 1$. I will use the following fact, commonly known as the  "path lifting property". If $\gamma$ is a path with such property, then, there exists a unique path $\tilde{\gamma}: [0, 1] \to \mathbb{R}$ such that $\tilde{\gamma}(0) = 0$ and $\gamma(t) = \mathrm{exp}(\mathrm{i}\tilde{\gamma}(t))$ for all $t$. Moreover, $\tilde{\gamma}$ has the same differentiability as $\gamma$. So we can take $\tilde{\gamma}_i$ two such path for $i = 0, 1$.
Since $\gamma_i(1) = 1$, $\tilde{\gamma}_i(1) = 2\pi n$ for some $n$. The claim is of course that this $n$ is actually what you called $k$, i.e that $n = \frac{1}{2\mathrm{i}\pi}\int_{\gamma_0}\frac{1}{z}\mathrm{d}z$. This can be computed explicitly since \begin{align*}
\int_{\gamma_0}\frac{1}{z}\mathrm{d}z &= \int_{0}^1\frac{\gamma_0'(t)}{\gamma_0(t)}dt\\
&= \int_0^1\frac{\mathrm{i}\tilde{\gamma}_0'(t)\mathrm{e}^{\mathrm{i}\tilde{\gamma}_0(t)}}{\mathrm{e}^{\mathrm{i}\tilde{\gamma}_0(t)}}dt\\
&= \mathrm{i(\tilde{\gamma}_0(1) - \tilde{\gamma}_0(0))}\\
&= 2\mathrm{i}\pi n
\end{align*}
And similarly for $\gamma_1$.
Now the problem becomes easier since we are reduced to finding an homotopy between $\tilde{\gamma}_0$ and $\tilde{\gamma}_1$ that leaves goth endpoint fixed. Since they both lie in a line, the most straightforward map $H(t, p) = p\tilde{\gamma}_0(t) + (1 - p)\tilde{\gamma}_1(t)$ will do. Composing this homotopy with $x \mapsto \mathrm{e}^{\mathrm{i}x}$ will give an homotopy between $\gamma_0$ and $\gamma_1$.
If you wish so, I can include a quick proof of the "path lifting lemma" I used.

Edit: As you requested, I will sketch a proof of the "path lifting lemma". Its exact statement is as follows: Let $\gamma: [0,1] \to S^1$, be a continuous path, and let $\theta_0 \in \mathbb{R}$ be such that $\gamma(0)=\mathrm{exp}(\mathrm{i}\theta_0)$. Then, there exists a unique continuous map $\tilde{\gamma}: [0,1] \to \mathbb{R}$ such that $\tilde{\gamma}(0) = \theta_0$ and, for all $t \in [0,1]$, $\gamma(t) = \mathrm{exp}(\mathrm{i}\tilde{\gamma}(t))$.
The proof is "piece by piece". $\tilde{\gamma}$ can be thought as a "continuous argument function" for $\gamma$. First, let's assume that $\gamma$ always have a real part strictly greater than $0$. Then, there is a well-defined formula for $\tilde{\gamma}$, namely, it is $\tan^{—1}\left(\frac{\mathrm{Im}(\gamma(t))}{\mathrm{Re}(\gamma(t))}\right) + 2k\pi$ where $k$ is so that the condition on $\theta_0$ is satisfied. This gives a continuous formula for $\tilde{\gamma}$, and is the only one possible with the extra condition on $\tilde{\gamma}(0)$. So we proved the lemma in this special case.
Now if $\gamma$ is such that its image is contained in any strict "half-circle", we can multiply gamma by some complex number lying on the circle to rotate it and to fall in the previous case, we only need to keep track on what it does to $\theta_0$ to keep the condition, but this pose no problem. So we can prove the lemma for any $\gamma$ that has a "small enough" image.
Finally, for any path $\gamma$, by compactness of $[0,1]$ and by continuity of $\gamma$, there are a finite number of points $t_0 = 0 < t_1 < \cdots < t_{n-1} < 1 = t_n$ such that $\gamma_{|[t_i,t_{i+1}]}$ has a "small enough" image in the previous sense. By a linear change of variable, we may apply successively the lemma, first to $\gamma_0 = \gamma_{|[0,t_1]}$ with $\tilde{\gamma}_0(0) = \theta_0$, then with $\gamma_1 = \gamma_{|[t_1,t_2]}$ with the condition $\tilde{\gamma}_1(t_1) = \tilde{\gamma}_0(t_0)$, and so on. We can then glue those maps to get a well-defined map $\tilde{\gamma}: [0,1] \to \mathbb{R}$. Uniqueness at each step shows uniqueness of $\tilde{\gamma}$. Note that all along we only used  inverse tangent, quotients and some rotations, so if $\gamma$ was $\mathcal{C}^k$, then so is $\tilde{\gamma}$.
