Solving a complex integral by parametrization and geometric series

I have to solve the given integral:

\begin{equation} \int_\gamma\frac{1}{z}+e^{z^2}dz \ \ \ \ \ where\ \ \gamma(t)=e^{-it} \ on \ t=[0,4\pi] \end{equation}

Since the first part is rapidly solved by parametrization, and the second is more difficult, I do the following on the first term (setting $z=e^{-it},\ \ dz=-ie^{-it}dt$):

\begin{equation} \int_\gamma\frac{1}{z}dz=-\int_0^{4\pi}\frac{1}{e^{-it}}ie^{-it}dt=-it\mid_0^{4\pi}=-4i\pi \end{equation}

The second however, I solve by writing $e^z\ ^{2}=\frac{1}{n!}\sum_{n=0}^{\infty}z^{(2+n)}$:

\begin{equation} \int_\gamma e^{z^2}dz = \frac{1}{n!}\int_\gamma \sum_{n=0}^{\infty}(z^2)^n dz=\frac{1}{n!}\int_\gamma \sum_{n=0}^{\infty}z^{2+n} dz \end{equation}

which gives by parameterizing $z=e^{-it}\ ,\ dz=-ie^{-it}dt$:

\begin{equation} \frac{1}{n!}\int_\gamma \sum_{n=0}^{\infty}z^{2+n} dz=-\frac{1}{n!}\int_0^{4\pi} \sum_{n=0}^{\infty}ie^{-it(2+n)}e^{-it}dt= -\frac{i}{n!}\int_0^{4\pi}\sum_{n=0}^{\infty}e^{-it(3+n)}dt=-\frac{i}{n!}\int_0^{4\pi}\sum_{n=0}^{\infty}e^{-it(3+n)}dt \end{equation}

which gives

\begin{equation} -\frac{i}{n!}\int_0^{4\pi}\sum_{n=0}^{\infty}e^{-it(3+n)}dt=-\frac{1}{n!(3+n)}\sum_{n=0}^{\infty}e^{-it(3+n)}\mid_0^{4\pi}=-\frac{1}{(n+3)n!}\sum_{n=0}^{\infty}\bigg(e^{-12i\pi(3+n)}-e^0\bigg) \end{equation}

which results in:

\begin{equation} -\frac{1}{(n+3)n!}\sum_{n=0}^{\infty}\bigg(e^{-12i\pi(3+n)}-e^0\bigg)=-\sum_{n=0}^{\infty}\frac{(-1 + e^{-4 i π(3 + n)}(3 + n))}{n!} \end{equation}

Taking the limit of this:

\begin{equation} \lim_{n\longrightarrow\infty}-\frac{(-1 + e^{-4 i π(3 + n)}(3 + n))}{n!}=0 \end{equation}

So going back to the original form:

\begin{equation} \int_\gamma\frac{1}{z}+e^{z^2}dz = -4\pi i +0 =-4\pi i \end{equation}

Would this be a correct approach to the original complex integral?

Thanks

$e^{z^2}$ is a holomorphic function. So, the integral along the closed curve $=0$.

Your approach for the first part is okay. But, by recognizing the curve is $|z|=1$ oriented clockwise for two rotations, you can immediately obtain $-4\pi i$ without calculation.