If $z_1=x_1+i y_1$, $z_2=x_2+i y_2$, $z_3=x_1+\frac{i x_2}{2}$, and $z_4=2 y_1+i y_2$ are complex numbers such that $|z_1|=1$, $|z_2|=2$, and $\operatorname{Re}(z_1 \bar{z}_2)=0$, then:

Let $z_1, z_2$ and $z_3$ be three complex numbers on the circle $|z|=1$ with $\arg(z_1) = \frac{-\pi}{4}, \arg(z_2) = 0$ and $\arg(z_3) = \frac{\pi}{4}$. If $|z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1|^2 = \alpha + \beta \sqrt{2}$,where $\alpha, \beta \in \mathbb{Z}$,then the value of $\alpha^2 + \beta^2$ is:

The expression $\frac{(1+i)^{n}}{(1-i)^{n-2}}$ equals

Let $z = a + ib, b \neq 0$ be a complex number satisfying $z^{2} = \overline{z} \cdot 2^{1-|z|}$. Then the least value of $n \in N$ such that $z^{n} = (z + 1)^{n}$ is equal to:

Among the statements:
$(S1) :$ The set $\{z \in \mathbb{C} - \{-i\} : |z|=1 \text{ and } \frac{z-i}{z+i} \text{ is purely real}\}$ contains exactly two elements,and
$(S2) :$ The set $\{z \in \mathbb{C} - \{-1\} : |z|=1 \text{ and } \frac{z-1}{z+1} \text{ is purely imaginary}\}$ contains infinitely many elements.

Let $S$ be the set of all $(\alpha, \beta)$ such that $\pi < \alpha, \beta < 2\pi$,for which the complex number $\frac{1-i \sin \alpha}{1+2i \sin \alpha}$ is purely imaginary and $\frac{1+i \cos \beta}{1-2i \cos \beta}$ is purely real. Let $Z_{\alpha \beta} = \sin 2\alpha + i \cos 2\beta$ for $(\alpha, \beta) \in S$. Then $\sum_{(\alpha, \beta) \in S} \left(i Z_{\alpha \beta} + \frac{1}{i \bar{Z}_{\alpha \beta}}\right)$ is equal to: