$\frac{(1+i)^{2016}}{(1-i)^{2014}}$ is equal to

If $a, b, c$ are non-zero real numbers with $c \neq 1$ such that $a^2+b^2+c^2=c$ and if $\alpha=\frac{a+i b}{1-c}$,then $a^2+b^2=$

If $i=\sqrt{-1}$,then $\sum_{n=0}^{\infty}\left(\frac{i}{3}\right)^n=$

Let $\alpha$ and $\beta$ be the roots of the equation $x^{2} + (2i - 1) = 0$. Then,the value of $|\alpha^{8} + \beta^{8}|$ is equal to

If $\cos \alpha+3 \cos 3 \beta+5 \cos 5 \gamma=0$,$\sin \alpha+3 \sin 3 \beta+5 \sin 5 \gamma=0$ and $\cos 3 \alpha+27 \cos 9 \beta+125 \cos 15 \gamma=\left(\lambda^2-4\right) \cos (\alpha+3 \beta+5 \gamma)$,then $\lambda$ is equal to

$\left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^8+\left(\frac{1+\cos \theta-i \sin \theta}{1+\cos \theta+i \sin \theta}\right)^{16}=$