Let C be the circle in the complex plane with centre z 0 = 1 2 1 + 3 i and radius r = 1 . Let z 1 = 1 + i…

Given,

$z_0=\frac{1+3i}{2}, z_1=\left(1+i\right)$

So, $\left|z_1-z_0\right|=\sqrt{{\left(1-\frac{1}{2}\right)}^2+{\left(1-\frac{3}{2}\right)}^2}=\sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{1}{\sqrt{2}}$

And $z_2$ be outside circle, $\left|z_1-z_0\right|\left|z_2-z_0\right|=1$

$\Rightarrow \frac{1}{\sqrt{2}}\left|z_2-z_0\right|=1$

$\Rightarrow \left|z_2-z_0\right|=\sqrt{2}$

Now given, $z_0, z_1 \& z_2$ are collinear,

So, by concept of rotation we get,

$\frac{z_2-z_0}{z_1-z_0}=\frac{\left|z_2-z_0\right|}{\left|z_1-z_0\right|}{e}^{i0}$

$\Rightarrow \frac{z_2-z_0}{z_1-z_0}=\frac{\left|z_2-z_0\right|}{\left|z_1-z_0\right|}\left(\pm 1\right)$

$\Rightarrow \frac{z_2-z_0}{z_1-z_0}=\pm 2$

$\Rightarrow z_2=z_0\pm 2\left(z_1-z_0\right)$

So, $z_2=2z_1-z_0=\frac{3}{2}+\frac{1}{2}i\Rightarrow {\left|z_2\right|}^2=\frac{5}{2}$

Or $z_2=3z_0-2z_1=\frac{-1}{2}+\frac{5}{2}i\Rightarrow {\left|z_2\right|}^2=\frac{13}{2}$

Hence, the smaller value will be, ${\left|z_2\right|}^2=\frac{5}{2}$