Prove-that-z-1-z-2-z-1-z-2-arg-z-1-arg-z-2-pi-2-

Question Number 19350 by Tinkutara last updated on 10/Aug/17

Prove that ∣ z_{1} + z_{2} ∣ = ∣ z_{1} - z_{2} ∣ \Leftrightarrow

\arg (z_{1}) - \arg (z_{2}) = \frac{π}{2}

Answered by ajfour last updated on 10/Aug/17

\Rightarrow ∣ z_{1} + z_{2} ∣^{2} =∣ z_{1} - z_{2} ∣^{2}

z_{1}^{2} + z_{2}^{2} + 2 Re (z_{1} {\bar{z}}_{2}) = z_{1}^{2} + z_{2}^{2} - 2 Re (z_{1} {\bar{z}}_{2})

\Rightarrow Re (z_{1} {\bar{z}}_{2}) = 0

Re [(x_{1} + {iy}_{1}) (x_{2} - {iy}_{2})] = 0

\Rightarrow x_{1} x_{2} = - y_{1} y_{2}

\Rightarrow \frac{y_{2}}{x_{2}} = - \frac{x_{1}}{y_{1}}

\Rightarrow \tan [\arg (z_{2})] = - \frac{1}{\tan [\arg (z_{1})]}

\Rightarrow \tan θ_{1} \tan θ_{2} + 1 = 0

and as \tan (θ_{1} - θ_{2}) = \frac{\tan θ_{1} - \tan θ_{2}}{1 + \tan θ_{1} \tan θ_{2}}

\Rightarrow θ_{1} - θ_{2} = \frac{π}{2}

\arg (z_{1}) - \arg (z_{2}) = \frac{π}{2} .

Commented by Tinkutara last updated on 10/Aug/17

Thank you very much Sir!