Suppose $f(z)=1+z+z^2+z^3.$ My question is
(1) How can we calculate $\int\limits_{[-1, i]}f(z)dz$
(2) How do we see $|\int\limits_{[-1, i]}(f(z))^ndz|\leq k(\frac{4}{3\sqrt{3}})^n$ where $k$ is a positive constant.
My idea:
(1) since $z$ is a complex number we can't just integrate it as usual.
(2) For this problem maybe we can use part (1) to prove it?
I'm not sure how to get a grip on this problem.
(1) In this case, you can integrate in the exact same way...
$$ \int_{[-1,i]} (1+z+z^2+z^3) dz = \left[z+\frac 12 z^2 + \frac 13 z^3 + \frac 14 z^4\right]_{-1}^i= \frac 13 + \frac{2i}{3} $$
you get the same answer as going through the computation of the line integral over the path $\gamma(t) = -1+t + ti, \quad t \in [0,1]$
$$ \int_{\Gamma} f(z) dz = \int_0^1 \gamma'(t) f(\gamma(t)) dt = \int_0^1 (1+i) f(-1+t+ti)dt = \frac 13 +\frac{2i}{3}. $$
[edit: comments on (2)]
The second part of the question seems trickier. What you can easily establish is that \begin{align*} \int_{[-1,i]} (1+z+z^2+z^3)^n dz= &\int_{[-1,i]}(1+z)^n(1+z^2)^n \\ = & \int_0^1 (1+i) (t(1+i)^n)(2(1-t)-2t(1-t)i)^ndt\\ =& (1+i)^{n+1} 2^n \int_0^1 t^n(1-t)^n(1-ti)^n dt \end{align*}
and so, $$ \left| \int_{[-1,i]} f(z)^n dz \right|\leq \sqrt{2} (2\sqrt{2})^n \int_0^1 t^n (1-t)^n \left(\sqrt{1+t^2}\right)^n dt $$
Although it is true that this bound is smaller than $\left(\frac{4}{3 \sqrt{3}}\right)^n$, and the result follows, I feel that there must be some simple trick to get the suggested bound.