Question

If arg ($Z_1$) = $\alpha$ , arg ($Z_2$) = $\beta$, the value of $|Z_1+Z_2{|}^2$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

We know that $|Z|{)}^2$ = z$\overline{z}$

$\Rightarrow$ $|Z_1+Z_2{|}^2$ = ($|z_1+z_2|$) $\overline{(}|z_1+z_2{|}^2)$

= $|Z_1+Z_2{|}^2$ $(\overline{z_1}+\overline{z_2})$ $[\because \overline{z_1+z_2}=\overline{z_1}+\overline{z_2}]$

= $z_1$ $\overline{z_1}$ + $z_1$ $\overline{z_2}$ + $z_2$ $\overline{z_1}$ + $z_2$ $\overline{z_2}$

= $|Z_1{|}^2$ + $|Z_2{|}^2$ + $z_1$ $\overline{z_2}$ + $z_2$ $\overline{z_1}$-------------------(1)

Let arg ($Z_1$) = $\alpha$, arg($Z_2$) = $\beta$

$z_1$ = $|Z_1|$ (cos $\alpha$ + i sin$\alpha$)

$z_2$ = $|Z_2|$ (cos $\beta$ + i sin$\beta$)

So, $z_1$ $\overline{z_2}$ + $z_2$ $\overline{z_1}$ = $|z_1|$.$|z_2|$.(cos $\alpha$ + i sin$\alpha$) (cos $\beta$ - i sin$\beta$) + $|z_1|$.$|z_2|$.(cos $\alpha$ - i sin$\alpha$) (cos $\beta$ + i sin$\beta$) $[\because arg(\overline{z})=-argz]$

= $|z_1|$.$|z_2|$ [cos$\alpha$.cos$\beta$ + i (-cos$\alpha$.sin$\beta$ + sin$\alpha$.cos$\beta$) + sin$\alpha$.sin$\beta$] + [cos$\alpha$.cos$\beta$ + i (cos$\alpha$.sin$\beta$ - sin$\alpha$.cos$\beta$) + sin$\alpha$.sin$\beta$]

= 2$|z_1|$.$|z_2|$ [cos$\alpha$.cos$\beta$ + sin$\alpha$.sin$\beta$] = 2$|z_1|$.$|z_2|$ = cos$(\alpha -\beta )$---------(2)

From (1) and (2),

$|Z_1+Z_2{|}^2$ = $|z_1{|}^2$ + $|z_2{|}^2$ + 2$|z_1|$.$|z_2|$ cos$(\alpha -\beta )$

Where $\alpha$ = arg$\left(z_1\right)$

$\beta$ = arg$\left(z_2\right)$