Let C be the circle in the complex plane with centre z _ 0 =...

Given,

${z}_{0} = \dfrac{1 + 3i}{2}, {z}_{1} = \left(1 + i\right)$

So, \left|{z}_{1} - {z}_{0}\right| = \sqrt{{\left(1 - \dfrac{1}{2}\right)}^{2} + {\left(1 - \dfrac{3}{2}\right)}^{2}} = \sqrt{\dfrac{1}{4} + \dfrac{1}{4}} = \dfrac{1}{\sqrt{2}}$</p><p>And${z}_{2}$be outside circle, \left|{z}_{1} - {z}_{0}\right|\left|{z}_{2} - {z}_{0}\right| = 1$

$⇒ \dfrac{1}{\sqrt{2}}\left|{z}_{2} - {z}_{0}\right| = 1$

$⇒ \left|{z}_{2} - {z}_{0}\right| = \sqrt{2}$

Now given, ${z}_{0}, {z}_{1} \text{ and } {z}_{2}$ are collinear,

So, by concept of rotation we get,

$\dfrac{{z}_{2} - {z}_{0}}{{z}_{1} - {z}_{0}} = \dfrac{\left|{z}_{2} - {z}_{0}\right|}{\left|{z}_{1} - {z}_{0}\right|}{e}^{i0}$

$⇒ \dfrac{{z}_{2} - {z}_{0}}{{z}_{1} - {z}_{0}} = \dfrac{\left|{z}_{2} - {z}_{0}\right|}{\left|{z}_{1} - {z}_{0}\right|}\left(\pm 1\right)$

$⇒ \dfrac{{z}_{2} - {z}_{0}}{{z}_{1} - {z}_{0}} = \pm 2$

$⇒ {z}_{2} = {z}_{0} \pm 2\left({z}_{1} - {z}_{0}\right)$

So, ${z}_{2} = 2{z}_{1} - {z}_{0} = \dfrac{3}{2} + \dfrac{1}{2}i ⇒ {\left|{z}_{2}\right|}^{2} = \dfrac{5}{2}$

Or ${z}_{2} = 3{z}_{0} - 2{z}_{1} = \dfrac{- 1}{2} + \dfrac{5}{2}i ⇒ {\left|{z}_{2}\right|}^{2} = \dfrac{13}{2}$

Hence, the smaller value will be, ${\left|{z}_{2}\right|}^{2} = \dfrac{5}{2}$