Could someone check if what I have done is correct?
To evaluate $\int_{C}\dfrac{z^3}{2z-i}$ where $C$ is the unit circle.
My solution is as follows:
Let $f(z):=\dfrac{z^3}{2z-i}$.
There seems to be a singularity at $z=\frac{i}{2}$, but it lies in the unit circle; Hence, I don't seem to be able to apply Cauchy-goursat, but I think it is a removable singularity, so I do the following.
Rewriting $f(z)$ as $f(z)=\dfrac{1}{2}{z^2}(1-i/2z)^{-1}$, Now we can expand this as a geometric series, thus we have $f(z)=\dfrac{1}{2}{z^2}[1-i/2z + (i/2z)^2+\dots]$
This function is complex differentible in the unit disc, hence by cauchy-goursat the integral over the unit circle is zero.
EDIT: I haven't yet reached Cauchy's Integral Formula, So am not allowed to use it yet.
Following Kavi Ramamurthy's idea:
Since $\int_{C}\dfrac{1}{z^n}=0$ for $n>1$ , and $\int_{C}z^n=0$ for $n\geq0$, we see that the integral simplifies to $\int_C \frac{-i}{16z}$ whose value is $\frac{\pi}{8}$.
Best Answer
If you integrate the series you obtained term by term you will find that all but one term integrates to $0$. The answer is $\int_C \frac {z^{2}} 2 (\frac i {2z})^{3} dz=\frac {\pi} 8$.