Fourier series without Fourier analysis techniques

It is known that one can sometimes derive certain Fourier series without alluding to the methods of Fourier analysis. It is often done using complex analysis. There is a way of deriving the formula
$$\sum_{k = 1}^\infty \frac{\sin(kz)}{k} = \frac{\pi - z}{2}$$
using complex analysis for some $z$. In other words, it can be shown that $\sum_{k = 1}^\infty \sin(kz)/k$ converges to $(\pi - z)/2$ for some $z$. My question is: How can we show that the formula above holds for $z \in (0, 2\pi)$ without using Fourier analysis?

Edit:

Using blue's suggestion I concocted the following proof.

Let $\log z$ be the principal value of the logarithm (with the branch cut along the negative real axis). Recall that $e^{iz} - e^{-iz} = 2i\sin z$ and $\text{Arg}(z) = \text{Arg}(z^*)$. Furthermore, we have
\begin{align*}
\log(1 - e^{iz}) &= -\sum_{k = 1}^\infty \frac{e^{ikz}}{k} \\
&= \log|1 - e^{iz}| + i\theta \\
&= \log\left|2\sin\frac{z}{2}\right| + i\theta,
\end{align*}
where $\theta = \text{Arg}(1 - e^{iz})$. Now, write $1 - e^{iz} = 1 - \cos z - i\sin z$ and let $z \in (0, \pi)$. Then
$$\tan \theta = \frac{\sin z}{1 - \cos z} = \frac{\cos(z/2)}{\sin(z/2)} = \tan\left(\frac{\pi}{2} - \frac{z}{2}\right).$$

Hence, $\theta = \pi/2 - z/2$. Moreover, since $z \in (0, \pi)$, we have $\theta \in (0, \pi/2)$.
Using Euler's formula $e^{iz} = \cos z + i \sin z$ and the fact that the sine function is odd, we see that equating the imaginary parts gives
$$\sum_{k = 1}^\infty \frac{\sin kz}{k} = \frac{\pi - z}{2}, \quad z \in (-\pi, 0) \cup (0, \pi).$$
Lastly, observe that $(-\pi, 0) \cup (0, \pi)$ can be replaced by $(0, 2\pi)$ due to the periodicity of the sine function.

Answer

In order to make sense of sines with complex arguments, you need to decompose them into complex exponentials. After that notice you're working with two series, each a Taylor expansion of the natural logarithm function (written appropriately). Proceed from there.