# |z1+z2|=|z1|+|z2| if and only if arg(z1)=arg(z2) , where z1 and z2 are complex numbers . how? please explain!

Nov 27, 2017

Kindly refer to the Discussion in the Explanation.

#### Explanation:

Let,

|z_j|=r_j; r_j gt 0 and arg(z_j)=theta_j in (-pi,pi]; (j=1,2).

$\therefore {z}_{j} = {r}_{j} \left(\cos {\theta}_{j} + i \sin {\theta}_{j}\right) , j = 1 , 2.$

Clearly, $\left({z}_{1} + {z}_{2}\right) = {r}_{1} \left(\cos {\theta}_{1} + i \sin {\theta}_{1}\right) + {r}_{2} \left(\cos {\theta}_{2} + i \sin {\theta}_{2}\right) ,$

$= \left({r}_{1} \cos {\theta}_{1} + {r}_{2} \cos {\theta}_{2}\right) + i \left({r}_{1} \sin {\theta}_{1} + {r}_{2} \sin {\theta}_{2}\right) .$

Recall that, $z = x + i y \Rightarrow | z {|}^{2} = {x}^{2} + {y}^{2.}$

$\therefore | \left({z}_{1} + {z}_{2}\right) {|}^{2} = {\left({r}_{1} \cos {\theta}_{1} + {r}_{2} \cos {\theta}_{2}\right)}^{2} + {\left({r}_{1} \sin {\theta}_{1} + {r}_{2} \sin {\theta}_{2}\right)}^{2} ,$

$= {r}_{1}^{2} \left({\cos}^{2} {\theta}_{1} + {\sin}^{2} {\theta}_{1}\right) + {r}_{2}^{2} \left({\cos}^{2} {\theta}_{2} + {\sin}^{2} {\theta}_{2}\right) + 2 {r}_{1} {r}_{2} \left(\cos {\theta}_{1} \cos {\theta}_{2} + \sin {\theta}_{1} \sin {\theta}_{2}\right) ,$

$= {r}_{1}^{2} + {r}_{2}^{2} + 2 {r}_{1} {r}_{2} \cos \left({\theta}_{1} - {\theta}_{2}\right) ,$

$\Rightarrow | {z}_{1} + {z}_{2} {|}^{2} = {r}_{1}^{2} + {r}_{2}^{2} + 2 {r}_{1} {r}_{2} \cos \left({\theta}_{1} - {\theta}_{2}\right) \ldots . \left({\star}^{1}\right)$.

$\text{Now Given that, } | {z}_{1} + {z}_{2} | = | {z}_{1} | + | {z}_{2} | ,$

$\iff | \left({z}_{1} + {z}_{2}\right) {|}^{2} = {\left(| {z}_{1} | + | {z}_{2} |\right)}^{2} = | {z}_{1} {|}^{2} + | {z}_{2} {|}^{2} + 2 | {z}_{1} | | {z}_{2} | , i . e . ,$.

$| \left({z}_{1} + {z}_{2}\right) {|}^{2} = {r}_{1}^{2} + {r}_{2}^{2} + 2 {r}_{1} {r}_{2.} \ldots \ldots \left({\star}^{2}\right) .$

From $\left({\star}^{1}\right) \mathmr{and} \left({\star}^{2}\right)$ we get,

$2 {r}_{1} {r}_{2} \cos \left({\theta}_{1} - {\theta}_{2}\right) = {r}_{1} {r}_{2.}$

$\text{Cancelling } {r}_{1} {r}_{2} > 0 , \cos \left({\theta}_{1} - {\theta}_{2}\right) = 1 = \cos 0.$

$\therefore \left({\theta}_{1} - {\theta}_{2}\right) = 2 k \pi \pm 0 , k \in \mathbb{Z} .$

$\text{But, } {\theta}_{1} , {\theta}_{2} \in \left(\pi , \pi\right] , {\theta}_{1} - {\theta}_{2} = 0 , \mathmr{and} ,$

${\theta}_{1} = {\theta}_{2} , \text{ giving, } a r g \left({z}_{1}\right) = a r g \left({z}_{2}\right) ,$ as desired!

Thus, we have shown that,

$| {z}_{1} + {z}_{2} | = | {z}_{1} | + | {z}_{2} | \Rightarrow a r g \left({z}_{1}\right) = a r g \left({z}_{2}\right) .$

The converse can be proved on similar lines.

Enjoy Maths.!