If for complex numbers z1 and z2,

Solution:

According to the question,

Let z1 = |z1| (cos θ1 + I sin θ1) and z2 = |z2| (cos θ2 + I sin θ2)

We have,

arg (z1) – arg (z2) = 0

⇒ θ1 – θ2 = 0

⇒ θ1 = θ2

We also have,

z2 = |z2| (cos θ1 + I sin θ1)

⇒ z1 – z2 = ((|z1|cos θ1 – |z2| cos θ1) + i (|z1| sin θ1 – |z2| sin θ1))

$\Rightarrow\left|\mathrm{z}_{1}-\mathrm{z}_{2}\right|=\sqrt{\left(\left|\mathrm{z}_{1}\right| \cos \theta_{1}-\left|\mathrm{z}_{2}\right| \cos \theta_{1}\right)^{2}+\left(\left|\mathrm{z}_{1}\right| \sin \theta_{1}-\left|\mathrm{z}_{2}\right| \sin \theta_{1}\right)^{2}}$

$=\sqrt{\left|\mathrm{z}_{1}\right|^{2}+\left|\mathrm{z}_{2}\right|^{2}-2\left|\mathrm{z}_{1}\right|\left|\mathrm{z}_{2}\right| \cos ^{2} \theta_{1}-2\left|\mathrm{z}_{1}\right|\left|\mathrm{z}_{2}\right| \sin ^{2} \theta_{1}}$

$=\sqrt{\left|\mathrm{z}_{1}\right|^{2}+\left|\mathrm{z}_{2}\right|^{2}-2\left|\mathrm{z}_{1}\right|\left|\mathrm{z}_{2}\right|\left[\cos ^{2} \theta_{1}+\sin ^{2} \theta_{1}\right]}$

We know that $\cos ^{2} \theta+\sin ^{2} \theta=1$

$=\sqrt{\left|\mathrm{z}_{1}\right|^{2}+\left|\mathrm{z}_{2}\right|^{2}-2\left|\mathrm{z}_{1}\right|\left|\mathrm{z}_{2}\right|}$

 

$=\sqrt{\left(\left|\mathrm{z}_{1}\right|-\left|\mathrm{z}_{2}\right|\right)^{2}}$

Hence, |z1 – z2| = |z1| – |z2|

Hence proved.