Let $\gamma : [0,1] \to \mathbb C$ be the contour given by $\gamma(t) = t + it$. Suppose $f : \mathbb C \to \mathbb C$ is a complex differentiable function which satisfies $|f(z)| \le 3$ for all $z$ such that $|z| \le 100$. Determine the maximum possible value of
$$\int_\gamma f(z)dz$$
My first approach was thinking of an $f(z)$ that satisfied the assumptions, say $f(z)=\frac{|z|}{100} + 3$ and use that, but I wasn't sure whether it would find the true maximum.
Note: I assume you want to find the maximum modulus.
$$ I_f=\int _\gamma f(z)dz=\int_0^1 f(\gamma(t))\gamma'(t) dt=\int_0^1f((1+i)t)(1+i) dt. $$ To find the maximum value of $|I_f|$, firstly note that $$ |I_f|\leq \int_0^1 |f((1+i)t)||1+i| dt=\sqrt 2 \int_0^1 |f((1+i)t)| dt\leq \sqrt 2\int_0^1 3dt=3\sqrt2. $$ Then notice that for the function $f(z)=3$ (for all $z$) we have $$|I_f|=|\int_0^1 3(1+i) dt|=3|1+i|=3\sqrt 2. $$ The maximum value of $|I_f|$ is therefore $3\sqrt 2$.