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

If $z \ne 0$ is a complex number,then

If $-i$ and $\alpha$ are the roots of the equation $iz^2 - 2(i+1)z + (2-i) = 0$,$\tan \theta = \frac{-1}{2}$ and $\theta \in 4^{\text{th}}$ quadrant,then $5^3 \cos 6\theta =$

Let $f(x) = x^4 + ax^3 + bx^2 + c$ be a polynomial with real coefficients such that $f(1) = -9$. Suppose that $i\sqrt{3}$ is a root of the equation $4x^3 + 3ax^2 + 2bx = 0$,where $i = \sqrt{-1}$. If $\alpha_1, \alpha_2, \alpha_3$,and $\alpha_4$ are all the roots of the equation $f(x) = 0$,then $|\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 + |\alpha_4|^2$ 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

If $z=1-\sqrt{3} i$,then $z^3-3 z^2+3 z=$