How to prove that
$$ (1+\cos \alpha +i\sin \alpha )^{100} = 2^{100}\left( \cos \left(\frac{\alpha}{2}\right)\right) ^{100} \left( \cos \left(\frac{100\alpha}{2}\right)+i\sin \left(\frac{100\alpha}{2}\right)\right)$$
I just need a hint. I tried to write $1+\cos \alpha +i\sin \alpha$ in polar form and use De,Moivre theorem. But it was impossible to compute $\arctan \frac{\sin \alpha}{1+\cos \alpha}$.
$$1+\cos\alpha+i\sin \alpha=1+e^{i\alpha}=e^{\frac{i\alpha}{2}}(e^{-\frac{i\alpha}{2}}+e^{\frac{i\alpha}{2}})=2\cos\left(\frac{\alpha}{2}\right)e^{\frac{i\alpha}{2}}$$
Hint: By double-angle formulas for sine and cosine, we have $$\frac{\sin\alpha}{1+\cos\alpha}=\frac{2\sin(\alpha/2)\cos(\alpha/2)}{2\cos^2(\alpha/2)}.$$
The figure below should make it easy to simplify the expression $\arctan \dfrac{\sin\alpha}{1+\cos\alpha}$. Note the isoceles triangle; you are looking for the angle at vertex $A$.
$$ \begin{aligned} & (1+\cos \alpha+i \sin \alpha)^{100} \\ = & \left(2 \cos ^2 \frac{\alpha}{2}+2 i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}\right)^{100} \\ = & {\left[2 \cos \frac{\alpha}{2}\left(\cos \frac{\alpha}{2}+i \sin \frac{\alpha}{2}\right)\right]^{100} } \\ = & 2^{100} \cos ^{100} \frac{\alpha}{2}(\cos 50 \alpha+i \sin 50 \alpha) \end{aligned} $$
