The modulus of the complex number $z$ such that $|z + 3 - i| = 1$ and $\arg{(z) = \pi}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
Let $z = x + iy$
$	herefore\ \ \ \ |z + 3 - i| = \left| (x + 3) + i(y - 1) ight| = 1$
$\Rightarrow \ \ \ \ \ \sqrt{(x + 3)^{2} + (y - 1)^{2}} = 1$ …(i)
$\because\ \ \ \arg{z = \pi\ \ \ \ \ \ \ \ \ \ \  \Rightarrow \ \ \ \ 	an^{- 1}{\frac{y}{x} = \pi}}$
$\Rightarrow \ \ \ \ \ \ \frac{y}{x} = 	an\pi = 0\ \ \ \ \$ $\Rightarrow \ \ y = 0\ \$ …(ii)
from Eqs.(i)and (ii)we get
$x = - 3,\ \ \ y = 0$
$	herefore\ \ \ z = - 3$
$\Rightarrow \ \ \ |z| = | - 3| = 3$

Q. The modulus of the complex number $z$ such that $|z + 3 - i| = 1$ and $\arg{(z) = \pi}$ is equal to

Explanation

Reviewed by Dr. Abubacker

Want more practice?