Question

Let $z_1,z_2$ be any two complex numbers and a, b two real numbers such that $a^2+b^2\ne 0$, then $|z_1{|}^2+|z_2{|}^2-|z_1^2+z_2^2|\le \frac{2|a{z}_1+b{z}_2{|}^2}{a^2+b^2}\le |z_1{|}^2+|z_2{|}^2+|z_1^2+z_2^2|$.
If this is true enter 1, else enter 0.

Answer 1

Let $z_1=r_1(cos{\theta }_1+isin{\theta }_2),z_2=r_2(cos{\theta }_2+isin{\theta }_2),a=rcos\alpha$ and $b=rsin\alpha$
then
$|z_1|=r_1,|z_2|=r_2$ and $a^2+b^2=r^2$ ........... (1)
Now $|a{z}_1+b{z}_2{|}^2=(a{z}_1+b{z}_2)(a\overline{z_1}+b\overline{z_2})$
$=a^2{z}_1\overline{z_1}+ab{z}_1\overline{z_2}+ab{z}_2\overline{z_1}+b^2{z}_2\overline{z_2}$
$=a^2|z_1{|}^2+b^2|z_2{|}^2+ab(z_1\overline{z_2}+\overline{z_1}{z}_2)$
$=a^2|z_1{|}^2+b^2|z_2{|}^2+2abRe\left(z_1\overline{z_2}\right)$
$=r^2{r}_1^{2}co{s}^2\alpha +r^2{r}_2^{2}si{n}^2\alpha +2r^{2}cos\alpha sin\alpha r_1{r}_{2}cos({\theta }_1-{\theta }_2)$
$=r^2\{r_1^{2}co{s}^2\alpha +r_2^{2}si{n}^2\alpha +2r_1{r}_{2}cos\alpha sin\alpha cos({\theta }_1-{\theta }_2\left)\right\}$
$=\frac{r^2}{2}\left\{r_1^2\right(1+cos2\alpha )+r_2^2(1-cos2\alpha )+2r_1{r}_{2}sin2\alpha cos({\theta }_1-{\theta }_2\left)\right\}$
$=\frac{r^2}{2}\{A+Bcos2\alpha +Csin2\alpha \}$
when $A=r_1^2+r_2^2,B=r_1^2-r_2^2$ and $C=2r_1{r}_{2}cos({\theta }_1-{\theta }_2)$
$\therefore \frac{2|a{z}_1+b{z}_2{|}^2}{r^2}=A+Bcos2\alpha +Csin2\alpha$
$\Rightarrow 2\frac{2|a{z}_1+b{z}_2{|}^2}{(a^2+b^2)}=A+Bcos2\alpha +Csin2\alpha$ {from (1)}
$=A+\sqrt{(B^2+C^2)}\{\frac{B}{\sqrt{B^2+C^2}}cos2\alpha +\frac{C}{\sqrt{B^2+C^2}}sin2\alpha \}$
Now let $B=Rsin\varphi$ and $C=Rcos\varphi$
$\therefore \frac{2|a{z}_1+b{z}_2{|}^2}{(a^2+b^2)}=A+Rsin(2\alpha +\varphi )$
Hence max. and min. values of $\frac{2|a{z}_1+b{z}_2{|}^2}{(a^2+b^2)}$ are respectively A + R and A - R
$\therefore \Rightarrow A-R\le \frac{2|a{z}_1+b{z}_2{|}^2}{a^2+b^2}\le A+R$ ........ (2)
$R=\sqrt{(B^2+C^2)}=\sqrt{\left\{\right(r_1^2-r_2^2{)}^2+4r_1^2{r}_2^{2}co{s}^2({\theta }_1-{\theta }_2)\}}$
$=\sqrt{\{r_1^4+r_2^4+2r_1^2{r}_2^{2}cos(2{\theta }_1-2{\theta }_2\left)\right\}}$