How does one define the complex distribution $1/z$?
Let $\phi(x)$ be a suitable test function ($\phi(x)\in C^\infty$ and has compact support on $(-\infty,\infty)$).
Then, write
$$\begin{align} \lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{1}{x+iy}\,dx&=\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{x-iy}{x^2+y^2}\,dx\\\\ &=\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{x}{x^2+y^2}\,dx-i\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{y}{x^2+y^2} \,dx\\\\ \end{align}$$
To evaluate the first limit, $\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{x}{x^2+y^2}\,dx$, we first write$$\begin{align}\int_{-\infty}^\infty \phi(x)\frac{x}{x^2+y^2}\,dx&=\int_{-\infty}^{-\epsilon}\phi(x)\frac{x}{x^2+y^2}\,dx\\\\&+\int_{-\epsilon}^\epsilon \phi(x)\frac{x}{x^2+y^2}\,dx\\\\&+\int_{\epsilon}^\infty \phi(x)\frac{x}{x^2+y^2}\,dx\tag 1\end{align}$$Inasmuch as $\phi$ is of compact support and $\phi(x)\frac{x}{x^2+y^2}$ is continuous, we have the limits$$\lim_{y\to 0^+}\int_{-\infty}^{-\epsilon} \phi(x)\frac{x}{x^2+y^2}\,dx=\int_{-\infty}^{-\epsilon} \frac{\phi(x)}{x}\,dx \tag2$$and$$\lim_{y\to 0^+}\int_{\epsilon}^\infty \phi(x)\frac{x}{x^2+y^2}\,dx=\int_{\epsilon}^\infty \frac{\phi(x)}{x}\,dx\tag3$$Next, we use Taylor's theorem to write $\phi(x)=\phi(0)+\phi'(0)x+o(x)$ for $x\in [-\epsilon,\epsilon]$.Then,$$\begin{align}\int_{-\epsilon}^\epsilon \phi(x)\frac{x}{x^2+y^2}\,dx&=\color{red}{\int_{-\epsilon}^\epsilon \phi(0)\frac{x}{x^2+y^2}\,dx}+\color{blue}{\int_{-\epsilon}^\epsilon \phi'(0)\frac{x^2}{x^2+y^2}\,dx}+\color{orange}{\int_{-\epsilon}^\epsilon \frac{o(x^2)}{x^2+y^2}\,dx}\\\\ &=\color{red}{0}+\color{blue}{\phi'(0)\left(2\epsilon +2y\arctan\left(\frac{\epsilon}{y}\right)\right)}+\color{orange}{o(\epsilon)}\tag 4 \end{align}$$Letting $y\to 0^+$ in $(4)$ we obtain$$\lim_{y\to 0^+}\int_{-\epsilon}^\epsilon \phi(x)\frac{x}{x^2+y^2}\,dx=2\phi'(0)\epsilon+o(\epsilon) \tag5 $$Finally using $(2)-(5)$, and letting $\epsilon\to 0^+$ yields$$\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{x}{x^2+y^2}\,dx=\lim_{\epsilon\to 0^+}\left(\int_{-\infty}^{-\epsilon} \frac{\phi(x)}{x}\,dx +\int_{\epsilon}^\infty \frac{\phi(x)}{x}\,dx \right)\equiv \text{PV}\left(\int_{-\infty}^\infty \frac{\phi(x)}{x}\,dx\right)$$
as was to be shown!
To evaluate the second limit, $\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{y}{x^2+y^2} \,dx$, we write
$$\begin{align} \lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{y}{x^2+y^2} \,dx&=\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(yx)\frac{1}{x^2+1} \,dx\tag6 \end{align}$$
whence applying the Dominated Convergence Theorem to $(6)$ yields the coveted result
$$\begin{align} \lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{y}{x^2+y^2} \,dx&=\int_{-\infty}^\infty \lim_{y\to 0^+}\left(\phi(yx)\right)\frac{1}{x^2+1} \,dx\\\\ &=\phi(0)\int_{-\infty}^\infty \frac{1}{x^2+1}\,dx\\\\ &=\pi \phi(0) \end{align}$$
as expected!
And we are done!
$\text{pv}.(\frac{1}{x})$ is easily understood as the distributional derivative of $\log |x|$, which means for any $\phi \in C^\infty$ with $\phi,\phi'$ decreasing faster than $1/x^2$ : $$\int_{-\infty}^\infty \phi(x) \text{pv}.(\frac{1}{x}) dx = -\int_{-\infty}^\infty \phi'(x) \log |x| dx$$ Now $$\log |x| = \log x- i \pi 1_{x < 0}= \lim_{y \to 0^+}\log (x+iy)- i \pi 1_{x < 0}$$ where the convergence is in $L^1_{loc}$. Therefore, as distributions $$\boxed{\text{pv}.(\frac{1}{x}) = \frac{d}{dx}\log |x| = \lim_{y \to 0^+} \frac{d}{dx}(\log (x+iy)- i \pi 1_{x < 0})= i \pi \delta(x)+\lim_{y \to 0^+}\frac{1}{x+iy}}$$