If $z$ is a point on the circle $|z|=1$ with $\operatorname{Arg}(z)=\frac{\pi}{6}$,then $\frac{z^{12}+1-z^6}{z^{12}+i z^6-1}=$

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:

If $(\cos \theta + i\sin \theta )(\cos 2\theta + i\sin 2\theta ) \dots (\cos n\theta + i\sin n\theta ) = 1$,then the value of $\theta$ is

Find the maximum value of $|z|$ when $\left|z-\frac{3}{z}\right|=2,$ where $z$ is a complex number.

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:

Let $z \in \mathbb{C}$ and $i=\sqrt{-1}$. If $a, b, c \in (0,1)$ are such that $a^2+b^2+c^2=1$ and $b+ic=(1+a)z$,then $\frac{1+iz}{1-iz}=$