I want to solve $\int_C Im(z) dz$ where $C$ is the unit circle. The first thing is to parameterize the curve, $C$: $$ z(t)=e^{it} $$ where $0\leq t \leq 2\pi$
Then $$ \frac {dz}{dt} = ie^{it} $$
and: $$ f(z(t)) = Im(z(t)) = Im(e^{it})=Im(\cos(t)+i\sin(t))=i\sin(t) $$
Then $\int_C f(z) dz$ = $\int\limits_{a}^{b} f(z(t))\cdot z'(t)dt = \int\limits_{0}^{2\pi} i\sin(t)\cdot ie^{it}dt = -\int\limits_{0}^{2\pi} \sin(t)\cdot e^{it} dt$
Let $I = \int\limits_{0}^{2\pi} \sin(t)\cdot e^{it} dt$, by integration by parts: $$ u=sin(t) $$ $$ du = cos(t)dt $$ $$ dv=e^{it}dt $$ $$ v=\frac{1}{i}e^{it} $$
And then $$ -I=\frac{1}{i}\sin(t)\cdot e^{it}-\frac{1}{i}\int\limits_{0}^{2\pi} \cos(t)\cdot e^{it} dt $$
By another integration by parts:
$$ u=\cos(t) $$ $$ du = -\sin(t)dt $$ $$ dv=e^{it}dt $$ $$ v=\frac{1}{i}e^{it} $$
At the end: $$ -I=\frac 1i\sin(t)\cdot e^{it}-\frac 1i(\frac 1i\cos(t)e^{it}+\frac 1iI) $$ $$ -I=\frac 1i\sin(t)\cdot e^{it}+\cos(t)e^{it}+I $$
$$ I=\frac {\frac 1i\sin(t)\cdot e^{it}+cos(t)\cdot e^{it}}{-2} $$
and if I substitude $2\pi$ and $0$: $$ -\frac 12-\frac 12=0 $$ But the book expect the answer be $-\pi$. What's wrong about my solution? Another question I have is that $Im(z)$ is analytic in everywhere and $C$ (the unit circle) is a closed path, so according to cauchy's integral theorem, the integral should be $0$, isn't that right?
It's difficult to integrate by parts here without making some small unnecessary mistakes. (For example, the left hand side after your first integration should be $I$, not $-I$.)
A simpler approach is:
\begin{align} \int_0^{2\pi} \operatorname{Im}(z(t)) z'(t)\,dt &= \int_0^{2\pi} \frac{e^{it}-e^{-it}}{2i} \cdot ie^{it} \,dt \\ &= \frac12\int_0^{2\pi} (e^{2it}-1)\,dt \\ &= \frac12 \left[ \frac{e^{2it}}{2i}-t \right]_0^{2\pi} = -\pi. \end{align}
As for your final question $ \operatorname{Im}(z)$ is in fact nowhere analytic.