Question

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

• A. 3
• B. 2
• C. 9
• D. 4

Answer 1

Answer: (A)

Let $z=x+iy$
$\therefore |z+3-i|=|(x+3)+i(y-1)|=1$
$\Rightarrow \sqrt{(x+3{)}^2+(y-1{)}^2}=1\dots (i)$
$\because \arg z=\pi$
$\Rightarrow {\tan }^{-1}\frac{y}{x}=\pi$
$\Rightarrow \frac{y}{x}=\tan \pi =0$
$\Rightarrow y=0\dots (ii)$
From equations (i) and (ii), we get
$x=-3,y=0\therefore z=-3$
$\Rightarrow |z|=|-3|=3$