I would like to solve the inequality $|\int_{\gamma} \frac{1}{z^2}dz|\leq 2$ where $\gamma$ is the line $[i,2+i]$. I thought about using the Cauchy theorem in closing the path between $i$ and $2+i$, but I don't know if this is allowed.
Is it possible to proceed in this manner? Otherwise, is anyone could give me a simple hint to solve it?
On
Just use the Fundamental Theorem of Calculus. If $F$ is an antiderivative of $f$ in a neighbourhood of $\gamma$, then $\int_\gamma f(z)\; dz = F(end (\gamma)) - F(start(\gamma))$.
On
Just use the definition of the contour integral $$\int_\gamma f(z)dz=\int_a^bf(\alpha(t))\alpha'(t)dt$$
Where $\alpha$ is a parametrization of $\gamma$, now our curve $\gamma$ is the line segment joining $i$ and $2+i$, we can parametrize this as $\alpha(t)=t+i$, $0\leq t \leq 2$.
Then the integral is $$ \int_0^2\frac 1 {(t+i)^2}dt=\frac {-1}{2+i}+\frac 1 {i} $$
I'm sure you can take it from here.
Hint: integral $\le$ (bound of $|$integrand$|$)(length of path).