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  4. The modulus of the complex number z such that |z+3-i|=1 and arg (z)=π is equal to

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

Complex Numbers and Quadratic Equations

Solution:

Let $z=x+i y$
$\therefore |z+3-i|=|(x+3)+i(y-1)|=1$
$\Rightarrow \sqrt{(x+3)^{2}+(y-1)^{2}}=1\, \dots(i)$
$\because \arg z=\pi$
$ \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
$x=-3, y=0 \therefore z=-3$
$\Rightarrow |z|=|-3|=3$