Show that $\int_{|z|=1}\,\,\,(1+z)^{\alpha} dz=0$ Consider the principal branch of $f(z)=(1+z)^{\alpha}$, where $\alpha$ is real. Show that for $\alpha > -1$
$\int_{|z|=1}\,\,\,(1+z)^{\alpha} dz=0$.
Now the integrand doesn't have an antiderivative on $|z|\leq 1$ so how the result is zero? Any help would be great.
 A: Consider a branch of $z\mapsto z^\alpha$ sending $z=r\exp(\text{i}\theta)$ to $r^\alpha\exp(\text{i}\alpha\theta)$ for $r> 0$ and $\theta\in\left(-\frac\pi2,+\frac\pi2\right)$.  Then, the curve $\gamma$ define by $\gamma(t)=1-\exp(\text{i}t)$ for $t\in(0,2\pi)$ lives within $\Omega:=\mathbb{C}\setminus(-\infty,0]$, whence $\gamma$ does not cross the branch cut.  Note that $z\mapsto z^\alpha$ has an antiderivative on $\Omega$, which is given by $z\mapsto \frac{1}{\alpha+1}z\cdot z^{\alpha}$.  Now,
$$\int_{\partial \mathbb{D}}\,(1+z)^\alpha\,\text{d}z=\int_\gamma\,z^\alpha\,\text{d}z=\lim_{\epsilon\to 0^+}\,\left(\frac{1}{\alpha+1} z\cdot z^{\alpha}\right)\Big|_{z=\gamma(\epsilon)}^{z=\gamma(2\pi-\epsilon)}=0\,.$$
A: This is an improper integral for $\alpha < 0$, as $f$ has a pole on the curve itself at $z = -1$. Avid19's substitution will work, though, provided that the improper integral $\int_0^2 u^\alpha du$ converges, which is easily proven for $\alpha > -1$.
Edit: since Avid19 deleted his post that I referred to, first he rewrote the integral as $\int_0^{2\pi} (1 + e^{i\theta})^\alpha d(e^{i\theta})$, then substituted $u = 1 + e^{i\theta}$. Because of the pole, this is equal to $\int_0^2 u^\alpha du + \int_2^0 u^\alpha du$, which is $0$ provided the integral converges.
