Question

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

• A. 1
• B. 2
• C. 9
• D. 4
• E. 3

Answer 1

Answer: (E)

Let $z=x+iy$
$\therefore$ $|z+3-i|=1$
$\Rightarrow$ $|x+iy+3-i|=1$
$\Rightarrow$ $|x+3+i(y-1)|=1$
$\Rightarrow$ ${(x+3)}^2+{(y-1)}^2=1$ ...(i) But ${\tan }^{-1}\frac{y}{x}=\pi$
$\Rightarrow$ $y=0$ ...(ii)
$\therefore$ From Eqs. (i) and (ii), we get ${(x+3)}^2=0$
$\Rightarrow$ $x=-3,-3$
$\therefore$ Required complex number is $x+iy=-3$
$\therefore$ $|x+iy|=|-3|=3$