$\lim_{R \to \infty} \int_{-R}^{R}\frac{1}{t-z_0}dt=$? We are given $z_0 \in \mathbb C - \mathbb R$ (meaning $z_0$ has an imaginary component. it is not real).
Show that if $Im(z_0)>0$ then $\lim_{R \to \infty} \frac{1}{2\pi i}\int_{-R}^{R}\frac{1}{t-z_0}dt=\frac{1}{2}$, and it is equal to $-\frac{1}{2}$ if $Im(z_0)<0$.
What I tried:
$\lim_{R \to \infty} \frac{1}{2\pi i}\int_{-R}^{R}\frac{1}{t-z_0}dt=\lim_{R \to \infty}\frac{1}{2\pi i}(\ln|t-z_0|)_{-R}^{R}=\lim_{R \to \infty}\frac{1}{2\pi i}(\ln|R-z_0|-\ln|-R-z_0|)=\lim_{R \to \infty}\frac{1}{2\pi i}\ln|\frac{R-z_0}{-R-z_0}|$
I am unsure where to go from here. Maybe define $z_0=a+bi$ and actually see what $|\frac{R-z_0}{-R-z_0}|$ is?
 A: We can carry out this integration without direct appeal to real analysis.  That is, we need not split the integral into real and imaginary parts.  Rather, we cut the plane to uniquely define the log function.

NOTE:
The choice of branch cut is not unique.  In fact, any contour that starts at $z_0$ and terminates at the point at infinity suffices here.

To that end, we cut the plane parallel to the real axis, starting at $z_0$ and ending at $(-\infty,\text{Im}(z_0))$.  Then, the integral of interest can be written 
$$\begin{align}
\int_{-R}^R \frac{1}{t-z_0}\,dt&=\log(R-z_0)-\log(-R-z_0)\\\\
&=\log \left(\frac{|R-z_0|}{|R+z_0|}\right)+i\left(\arg(R-z_0)-\arg(-R-z_0)\right) \tag 1
\end{align}$$
where $-\pi < \arg(z-z_0)\le \pi$.  In arriving at $(1)$, we used 
$$\log (z-z_0) =\log |z-z_0|+i\arg(z-z_0)$$
For $\text{Im}(z_0)>0$, $\lim_{R\to \infty} \arg(R-z_0)=0$ and $\lim_{R\to \infty} \arg(-R-z_0)=-\pi$.  
For $\text{Im}(z_0)<0$, $\lim_{R\to \infty} \arg(R-z_0)=0$ and $\lim_{R\to \infty} \arg(-R-z_0)=\pi$.  
Putting it all together, we have
$$\frac{1}{2\pi i}\lim_{R\to \infty}\int_{-R}^{R}\frac{1}{t-z_0}\,dt=
\begin{cases}
\frac12 &,  \text{Im}(z_0)>0\\\\
-\frac12 &, \text{Im}(z_0)<0
\end{cases}
$$
as was to be shown!
A: I suppose $z_0=iy$ you can use a change of variables to obtain that,
$\int_{-R}^R{1\over{t-iy}}dt=\int_{-R}^R{{t+iy}\over{t^2+y^2}}dt$
$\int_{-R}^R{t\over{t^2+y^2}}dt =0$ since if $f(t)={t\over{t^2+y^2}}$, $f(-t)=-f(t)$
$\int_{-R}^R{y\over{t^2+y^2}}dt= Artan(R/y)-Arctan(-R/y)$ this implies that $lim_{R\rightarrow +\infty}\int_{-R}^R{y\over{t^2+y^2}}dt=\pi$ so $\int_{-R}^R{1\over{t-iy}}dt=1/2$.
A: If $z_0=a+bi$, then
$$
\frac{1}{t-z_0}=\frac{1}{t-a-bi}=\frac{t-a+bi}{(t-a)^2+b^2}
$$
Now
$$
\int_{-R}^R\frac{t-a}{(t-a)^2+b^2}\,dt=
\left[\frac{1}{2}\log((t-a)^2+b^2)\right]_{-R}^R
$$
and the limit of this is $0$.
Thus you want to compute
$$
\int_{-\infty}^\infty\frac{b}{(t-a)^2+b^2}\,dt=
\int_{-\infty}^{\infty}\frac{1}{u^2+1}\,du
$$
with the substitution $t-a=bu$. Note that the improper integral is not the limit of the integrals from $-R$ to $R$, but it exists nonetheless, so its value is the same as the limit you're looking for.
$$
