Calculate $\int_\alpha e^{z^2+z+1}+e^{Im(z)} \ dz $

Question

Calculate $$\int_\alpha e^{z^2+z+1}+e^{Im(z)} \ dz $$

$\alpha$ is the square of vertices: $$0,1,i,i+1$$

Different segments: $$\alpha_1,\alpha_2,\alpha_3,\alpha_4$$

$\alpha_i: [0,1] \rightarrow\mathbb{C}$

$$\alpha_1(t)=t$$ $$\alpha_2(t)=1+it$$ $$\alpha_3(t)=t+i$$ $$\alpha_4(t)=it$$

$$\alpha=\alpha_1+\alpha_2-\alpha_3-\alpha_4$$

$$\int_0^1 f(\alpha(t)) \ \alpha'(t) \ dt$$

I don't know how to calculate $f(\alpha(t))$ in this case: $e^{Im(z)}$.

$$\int_{\alpha_1} e^{z^2+z+1}+e^{Im(z)} \ dz=\int_0^1 e^{t^2+t+1}+e^{t} \ dt ?$$

Could I have any help, please?

Thanks!

Answer 1

We have \begin{align} \alpha_1(t)&=t\\ \alpha_2(t)&=1+ti\\ \alpha_3(t)&=1-t+i\\ \alpha_4(t)&=(1-t)i\\ &\text{where $0\leq t\leq1$} \end{align}

We have to evaluate $\int_\alpha (e^{z^2+z+1}+e^{Im(z)} )\ dz=\int_\alpha e^{Im(z)} \ dz$,since $e^{z^2+z+1}$ is analytic inside the rectangle.

Now,\begin{align} \int_\alpha e^{Im(z)} \ dz&=\int_{\alpha_1} e^{Im(z)} \ dz+\int_{\alpha _2} e^{Im(z)} \ dz+\int_{\alpha_ 3} e^{Im(z)} \ dz+\int_{\alpha_ 4} e^{Im(z)} \ dz\\ &=\int_0^1e^0dt+\int_0^1ie^tdt+\int_0^1-e dt+\int_0^1-ie^{1-t}dt\\ &=1+i(e-1)+(-e)+(-i)(e-1)\\ &=1-e \end{align}

Answer 2

By Cauchy:

$\int_\alpha e^{z^2+z+1} \ dz=0$.

Hence you only have to compute the integral

$\int_\alpha e^{Im(z)} \ dz$.