Let $\arg(z)$ represent the principal argument of the complex number $z$. The curves $|z|=3$ and $\arg(z-1)-\arg(z+1)=\frac{\pi}{4}$ intersect:

The locus of $z=x+iy$, such that $\operatorname{Im}\left(\frac{z-3i}{iz+4}\right)=0$ is

If $\frac{z-1}{z+1}$ is purely imaginary, then

Let $\theta_1, \theta_2, \ldots, \theta_{10}$ be positive valued angles (in radian) such that $\theta_1+\theta_2+\ldots+\theta_{10}=2 \pi$. Define the complex numbers $z_1=e^{i \theta_1}, z_k=z_{k-1} e^{i \theta_k}$ for $k=2,3, \ldots, 10$,where $i=\sqrt{-1}$. Consider the statements $P$ and $Q$ given below:
$P: |z_2-z_1|+|z_3-z_2|+\ldots+|z_{10}-z_9|+|z_1-z_{10}| \leq 2 \pi$
$Q: |z_2^2-z_1^2|+|z_3^2-z_2^2|+\ldots+|z_{10}^2-z_9^2|+|z_1^2-z_{10}^2| \leq 4 \pi$
Then,

Let $A = \{z \in \mathbb{C} : |z - 2| \le 4\}$ and $B = \{z \in \mathbb{C} : |z - 2| + |z + 2| = 5\}$. Then the maximum value of $\{|z_1 - z_2| : z_1 \in A \text{ and } z_2 \in B\}$ is

The equation $\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}$ represents a circle with: