Mathematics - Complex Number Question with Solution | TestHub

Let $z=x+iy$ Given, $|z+3-i|=1$ $\begin{array}{rl}& \Rightarrow |(x+3)+(y-1)i|=1\\ & \Rightarrow \sqrt{(x+3{)}^2+(y-1{)}^2}=1\\ & \Rightarrow (x+3{)}^2+(y-1{)}^2=1\dots (i)\end{array}$ Above is the equation of circle with centre $\equiv (-3,1)$ and $\text{ radius }=1$ Also, $\arg (z)=\pi \Rightarrow {\tan }^{-1}\left|\frac{y}{x}\right|=\pi$ $\Rightarrow \left|\frac{y}{x}\right|=\tan \pi =0\Rightarrow y=0$By Eq. (i), $\begin{array}{rlr}\Rightarrow & x& =-3\\ \therefore & z& =-3+0i\\ \Rightarrow & \text{ Modulus of }z& =3\end{array}$