# by using ML- inequality

• Aug 5th 2009, 12:32 AM
flower3
by using ML- inequality
prove that :
$\mid \int \frac{1}{{z^4}} dz \mid \leq 4 \sqrt 2$ on c where C: the line segment from z=i to z=1
• Aug 5th 2009, 02:36 AM
Random Variable
$\Big|\int_{C} \frac{1}{z^{4}} \ dz\Big| \le ML$

where M is an upper bound of $\Big|\frac{1}{z^{4}}\Big|$ on C and L is the length of C.

the length of C is $\sqrt{2}$

$\Big|\frac{1}{z^{4}}\Big| = \frac{1}{|z|^{4}}$

where z is the distance from a point on C to the origin

To maximize $\frac{1}{|z|^{4}}$ we have to minimize $|z|^{4}$

The shortest distance from a point on C to the origin is $\frac{1}{\sqrt{2}}$. (It's the length of the perpendicular bisector of C.)

so $|z^{4}| \ge \Big(\frac{1}{\sqrt{2}}\Big)^4 = \frac{1}{4}$

which implies that $\frac{1}{|z|^{4}} \le 4$

then $\int_{C} \frac{1}{z^{4}} \ dz \le 4 \sqrt{2}$