Definite Integrals #4 : It's going above integrals!

Method 1

Cauchy Integral Formula states that, $\oint_\gamma\dfrac{f(\zeta+z_{const})}{\zeta-z} d\zeta = 2\pi i\cdot f(z+z_{const})$

Also using, $\oint_\gamma F(z)~dz = \int_a^b F(\gamma(t))\cdot \dfrac{d\gamma}{dt}(t)~dt$

Taking $\gamma$ as the equation of a circle with radius $r$ and centre $z$ , also using $\zeta = re^{i\theta}$ , we get, $\begin{aligned}I &= \int_0^{4\pi} \left(z_0 + re^{i\theta}\right)^jd\theta,~~\color{#3D99F6}{\text{Taking }f(z) = z^j}\\ &=\dfrac2i\oint_\gamma\dfrac{f(\zeta+z_0)}{\zeta} d\zeta\\&=\dfrac2i\times 2\pi i\cdot f(z_0+0)\\&=4\pi f(z_0)=4\pi z_0^j\end{aligned}$

$\Huge \therefore \int_0^{4\pi} \left(z_0 + re^{i\theta}\right)^jd\theta =\boxed{4\pi z_0^j}$

Method 2

Using Binomial Expansion $\begin{aligned}I &= \int_0^{4\pi} \left(z_0 + re^{i\theta}\right)^jd\theta \\&=\int_0^{4\pi}\left(z_0^j + \binom j1z_0^{j-1}e^{i\theta} + \binom j2z_0^{j-2}e^{2i\theta} +\cdots\right)d\theta\end{aligned}$

Now, take a general term $\int_0^{4\pi}ce^{ni\theta}d\theta = \left.\dfrac{ce^{in\theta}}{in}\right|_0^{4\pi} = 0$ when $n\ne0$ . So, $\begin{aligned}I &= \int_0^{4\pi} \left(z_0 + re^{i\theta}\right)^jd\theta \\&=\int_0^{4\pi}\left(z_0^j + \binom j1z_0^{j-1}e^{i\theta} + \binom j2z_0^{j-2}e^{2i\theta} +\cdots\right)d\theta\\&=\int_0^{4\pi}z_0^j\;d\theta\\&=4\pi z_0^j\end{aligned}$

$\Huge \therefore \int_0^{4\pi} \left(z_0 + re^{i\theta}\right)^jd\theta =\boxed{4\pi z_0^j}$