prove that :
$\displaystyle \mid \int \frac{1}{{z^4}} dz \mid \leq 4 \sqrt 2 $ on c where C: the line segment from z=i to z=1
$\displaystyle \Big|\int_{C} \frac{1}{z^{4}} \ dz\Big| \le ML $
where M is an upper bound of $\displaystyle \Big|\frac{1}{z^{4}}\Big| $ on C and L is the length of C.
the length of C is $\displaystyle \sqrt{2} $
$\displaystyle \Big|\frac{1}{z^{4}}\Big| = \frac{1}{|z|^{4}} $
where z is the distance from a point on C to the origin
To maximize $\displaystyle \frac{1}{|z|^{4}} $ we have to minimize $\displaystyle |z|^{4} $
The shortest distance from a point on C to the origin is $\displaystyle \frac{1}{\sqrt{2}} $. (It's the length of the perpendicular bisector of C.)
so $\displaystyle |z^{4}| \ge \Big(\frac{1}{\sqrt{2}}\Big)^4 = \frac{1}{4} $
which implies that $\displaystyle \frac{1}{|z|^{4}} \le 4 $
then $\displaystyle \int_{C} \frac{1}{z^{4}} \ dz \le 4 \sqrt{2} $