Let C be the circle in the complex plane with centre z_0= 1 2(1+3i) and radius r=1. Let z_1=1+i and the complex number z_2 be outside circle C such…

Given,

$${z}_{0}=\frac{1+3i}{2},{z}_{1}=(1+i)$$z0​=21+3i​,z1​=(1+i)

So, $|{z}_{1}-{z}_{0}|=\sqrt{{(1-\frac{1}{2})}^{2}+{(1-\frac{3}{2})}^{2}}=\sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{1}{\sqrt{2}}$∣z1​−z0​∣=(1−21​)2+(1−23​)2​=41​+41​​=2​1​

And ${z}_{2}$z2​ be outside circle, $|{z}_{1}-{z}_{0}||{z}_{2}-{z}_{0}|=1$∣z1​−z0​∣∣z2​−z0​∣=1

$$\Rightarrow \frac{1}{\sqrt{2}}|{z}_{2}-{z}_{0}|=1$$⇒2​1​∣z2​−z0​∣=1

$$\Rightarrow |{z}_{2}-{z}_{0}|=\sqrt{2}$$⇒∣z2​−z0​∣=2​

Now given, {z}_{0},{z}_{1}&{z}_{2} are collinear,

So, by concept of rotation we get,

$$\frac{{z}_{2}-{z}_{0}}{{z}_{1}-{z}_{0}}=\frac{|{z}_{2}-{z}_{0}|}{|{z}_{1}-{z}_{0}|}{e}^{i0}$$z1​−z0​z2​−z0​​=∣z1​−z0​∣∣z2​−z0​∣​ei0

$$\Rightarrow \frac{{z}_{2}-{z}_{0}}{{z}_{1}-{z}_{0}}=\frac{|{z}_{2}-{z}_{0}|}{|{z}_{1}-{z}_{0}|}(\pm 1)$$⇒z1​−z0​z2​−z0​​=∣z1​−z0​∣∣z2​−z0​∣​(±1)

$$\Rightarrow \frac{{z}_{2}-{z}_{0}}{{z}_{1}-{z}_{0}}=\pm 2$$⇒z1​−z0​z2​−z0​​=±2

$$\Rightarrow {z}_{2}={z}_{0}\pm 2({z}_{1}-{z}_{0})$$⇒z2​=z0​±2(z1​−z0​)

So, ${z}_{2}=2{z}_{1}-{z}_{0}=\frac{3}{2}+\frac{1}{2}i\Rightarrow {|{z}_{2}|}^{2}=\frac{5}{2}$z2​=2z1​−z0​=23​+21​i⇒∣z2​∣2=25​

Or ${z}_{2}=3{z}_{0}-2{z}_{1}=\frac{-1}{2}+\frac{5}{2}i\Rightarrow {|{z}_{2}|}^{2}=\frac{13}{2}$z2​=3z0​−2z1​=2−1​+25​i⇒∣z2​∣2=213​

Hence, the smaller value will be, ${|{z}_{2}|}^{2}=\frac{5}{2}$∣z2​∣2=25​