Question

If $\left|z\right|\le 1,\left|\omega \right|\le 1,$ then ${|z-\omega |}^2$

• A. $\le {(\left|z\right|-\left|\omega \right|)}^2-{(arg\left(z\right)-arg\left(\omega \right))}^2$
• B. $\le {(\left|z\right|-\left|\omega \right|)}^2+{(arg\left(z\right)-arg\left(\omega \right))}^2$
• C. $\le {(\left|z\right|-\left|\omega \right|)}^2+2{(arg\left(z\right)-arg\left(\omega \right))}^2$
• D. None of these

Answer 1

The correct option is B $\le {(\left|z\right|-\left|\omega \right|)}^2+{(arg\left(z\right)-arg\left(\omega \right))}^2$

Let $z=r_1(\cos \alpha +i\sin \alpha )$

and, $\omega =r_2(\cos \beta +i\sin \beta )\Rightarrow arg\left(z\right)=\alpha ,arg\left(\omega \right)=\beta$

$\left|z\right|=r_1,\left|\omega \right|=r_2.$

Given: $\left|z\right|\le 1,\left|\omega \right|\le 1\Rightarrow r_1\le 1,r_2\le 1$

Now, consider ${|z-\omega |}^2={|(r_1\cos \alpha -r_2\cos \beta )+i(r_1\sin \alpha -r_2\sin \beta )|}^2$

$={(r_1\cos \alpha -r_2\cos \beta )}^2+{(r_1\sin \alpha -r_2\sin \beta )}^2={r_1}^2({\cos }^2\alpha +{\sin }^2\alpha )+{r_2}^2({\cos }^2\beta +{\sin }^2\beta )-2r_1{r}_2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )={r_1}^2+{r_2}^2-2r_1{r}_2\cos (\alpha -\beta )={(r_1-r_2)}^2+2r_1{r}_2(1-\cos (\alpha -\beta ))$

$={(r_1-r_2)}^2+4r_1{r}_2{\sin }^2\left(\frac{\alpha -\beta }{2}\right)\le {(r_1-r_2)}^2+\mathrm{4.1.1.}{\sin }^2\left(\frac{\alpha -\beta }{2}\right)\Rightarrow {|z-\omega |}^2\le {(\left|z\right|-\left|\omega \right|)}^2+{(arg\left(z\right)-arg\left(w\right))}^2$