complex integration on closed c

Question

Let $c: |z-i|=1$
evaluate integral $I=\int_c e^{z^2}+z^4+\frac{z^3+4z}{2z-i}+\frac{1}{z^2+1} \,dz$

this is my answer please correct me if I m wrong. I m learning.
$I=\int_c e^{z^2}dz\ +\int_cz^4dz\ +\int_c\frac{z^3+4z}{2z-i}dz\ +\int_c\frac{1}{z^2+1} \,dz$
since $e^{z^2}$,$z^4$ are analytic in and on c, then by Cauchy Goursat theorem, their integrals equal zero.
$\int_c\frac{1}{z^2+1}=\int_c\frac{1}{(z-i)(z+i)}=\pi$ . ( by Cauchy integral)
for $\int_c\frac{z^3+4z}{2z-i}dz\ $ , I since the order of numerator is more than the order of denominator, I did not say the root $z=i/2$ is inside c, so I have to use residue theorem. instead, I guess I need to divide as $\int_c\ [(2z-i) (\frac{z^2}{2}+\frac{zi}{2})]+\frac{7z}{4}dz\ $ and since now after division, the integrand in analytic inc, then integral is also zero,
so $I=\pi$
am I correct?

Answer 1

Yes, $\int_Ce^{z^2}\,\mathrm dz=\int_Cz^4\,\mathrm dz=0$.

And $\int_C\frac{\mathrm dz}{z^2+1}=\pi$.

But\begin{align}\int_C\frac{z^3+4z}{2z-i}\,\mathrm dz&=\int_C\frac{\frac12z^3+2z}{z-\frac i2}\\&=2\pi i\times\left(\frac12{\left(\frac i2\right)^3+2\frac i2}\right)\\&=-\frac{15\pi}{16}.\end{align}

Answer 2

For the last integral, forget about the order:

$\displaystyle \int_C \frac{z^3 + 4z}{2z - i} \, dz = \frac{1}{2} \int_C \frac{z^3 + 4z}{z - i/2} \, dz = \frac{1}{2} 2\pi i f(i/2)$ by the Cauchy formal where $f(z) = z^3 + 4z$.