If $z_{1}$ and $z_{2}$ are two non-zero complex numbers such that $∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣$ , then $a r g z_{1} - a r g z_{2}$ is equal to

Question

Verified by Toppr

Answer 1

Given $z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣$
$\Rightarrow ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} + 2 R e (z_{1} \overset{z_{2}}{ˉ}) = ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} + 2 ∣ z_{1} ∣ ∣ z_{2} ∣$
$\Rightarrow R e (z_{1} \overset{z_{2}}{ˉ}) = ∣ z_{1} ∣ ∣ z_{2} ∣$
$\Rightarrow R e (z_{1} \overset{z_{2}}{ˉ}) = ∣ z_{1} \overset{z_{2}}{ˉ} ∣$
(or $R e (\overset{z_{1}}{ˉ} z_{2}) = ∣ \overset{z_{1}}{ˉ} z_{2} ∣)$
$\Rightarrow z_{1} \overset{z_{2}}{ˉ} = R^{+} \Rightarrow A r g (z_{1} \overset{z_{2}}{ˉ}) = A r g (\overset{z_{1}}{ˉ} z_{2}) = 0$
$\Rightarrow A r g z_{1} + A r g \overset{z_{2}}{ˉ} = A r g \overset{z_{1}}{ˉ} + A r g z_{2} = 0$
$\Rightarrow A r g z_{1} - A r g z_{2} = - A r g z_{1} + A r g z_{2} = 0$

Answer 2

Given

z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣

\Rightarrow ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} + 2 R e (z_{1} \overset{z_{2}}{ˉ}) = ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} + 2 ∣ z_{1} ∣ ∣ z_{2} ∣

\Rightarrow R e (z_{1} \overset{z_{2}}{ˉ}) = ∣ z_{1} ∣ ∣ z_{2} ∣

\Rightarrow R e (z_{1} \overset{z_{2}}{ˉ}) = ∣ z_{1} \overset{z_{2}}{ˉ} ∣

(or

R e (\overset{z_{1}}{ˉ} z_{2}) = ∣ \overset{z_{1}}{ˉ} z_{2} ∣)

\Rightarrow z_{1} \overset{z_{2}}{ˉ} = R^{+} \Rightarrow A r g (z_{1} \overset{z_{2}}{ˉ}) = A r g (\overset{z_{1}}{ˉ} z_{2}) = 0

\Rightarrow A r g z_{1} + A r g \overset{z_{2}}{ˉ} = A r g \overset{z_{1}}{ˉ} + A r g z_{2} = 0

\Rightarrow A r g z_{1} - A r g z_{2} = - A r g z_{1} + A r g z_{2} = 0