Show $\left|\int_{|z - R| = R} z^{14}\, dz \right| \leq 2^{15} \pi R^{15}$

Question

$$\left|\int_{|z - R| = R} z^{14}\, dz \right| \leq 2^{15} \pi R^{15}$$

I'm not quite sure how to show this is true. I was using the formula:

$|\int f(z) dz|≤ \int|f(z)||dz|≤ ML$ where $f(z)$ was bounded so $|f(z)|≤ M$ and $L$ is length of the curve we're integrating over.

I tried taking the modulus of $f(z)$ and then $dz$, but that's where I got stuck.

Answer 1

Using $w=z-R, dw=dz$ the integral becomes:

$I=\int_{|z - R| = R} z^{14}dz =\int_{|w| = R} (w+R)^{14}dw=\int_{0}^{2\pi}R^{14}(e^{it}+1)^{14}iRe^{it}dt$

when using $w=Re^{it}, dw=iRe^{it}$, so:

$|I| \le R^{15}\int_{0}^{2\pi}|(e^{it}+1)^{14}ie^{it}|dt \le R^{15}2^{14} 2\pi=2^{15}\pi R^{15}$

(since $|(e^{it}+1)^{14}ie^{it}| \le 2^{14}$ as $|e^{it}+1| \le 2$), so done!

Answer 2

This could be done also without substitution. The maximal value that $|z|$ can take on the circle $\left \{ z|\ |z-R|<R \right \}$ is $2R$, the furthest point from the origin. So, the maximal value $|z|^{14}$ can take is $(2R)^{14}$. Obviously, $L=2\pi R$, so we get $$\left | \int_{\Gamma}z^{14}\text{d}z \right |\leq (2R)^{14}\cdot2\pi R$$