If $z = x + iy$ and the point $P$ represents $z$ in the Argand plane,then the locus of $z$ satisfying the equation $|z - 1| + |z + i| = 2$ is

The equation $|z+1-i|=|z-1+i|$ represents a (where $z$ is a complex number)

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,

The locus of $z$ such that $\left|\frac{z-i}{z+i}\right|=2$,where $z=x+iy$,is

If $P(x)=0$ is a polynomial equation of least degree with integer coefficients and $\sqrt{2}+\sqrt{3} i$ is one of its roots,then that equation is

If at least one value of the complex number $z = x + iy$ satisfies the condition $|z + \sqrt{2}| = a^2 - 3a + 2$ and the inequality $|z + i\sqrt{2}| < a^2$,then