Prove-that-z-1-z-2-2-z-1-2-z-2-2-z-1-z-2-is-purely-imaginary-number-

Question Number 19351 by Tinkutara last updated on 10/Aug/17

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

\frac{z_{1}}{z_{2}} is purely imaginary number .

Commented by dioph last updated on 10/Aug/17

z_{1} = a_{1} + {i b}_{1}, z_{2} = a_{2} + {i b}_{2}

∣ z_{1} + z_{2} ∣^{2} = ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} \Leftrightarrow

{(a_{1} + a_{2})}^{2} + {(b_{1} + b_{2})}^{2} = (a_{1}^{2} + b_{1}^{2}) + {(a_{2}^{2} + b_{2})}^{2} \Leftrightarrow

2 a_{1} a_{2} + 2 b_{1} b_{2} = 0 \Leftrightarrow

\frac{z_{1}}{z_{2}} = \frac{a_{1} a_{2} + b_{1} b_{2} + i (a_{2} b_{1} - a_{1} b_{2})}{a_{2}^{2} + b_{2}^{2}} =

= i \frac{a_{2} b_{1} - a_{1} b_{2}}{a_{2}^{2} + b_{2}^{2}}

Commented by Tinkutara last updated on 10/Aug/17

Thank you very much Sir!

Answered by ajfour last updated on 10/Aug/17

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

\Rightarrow 1 + ∣ \frac{z_{1}}{z_{2}} ∣^{2} + 2 Re (\frac{z_{1}}{z_{2}}) = 1 + ∣ \frac{z_{1}}{z_{2}} ∣^{2}

or Re (\frac{z_{1}}{z_{2}}) = 0

\Rightarrow \frac{z_{1}}{z_{2}} is purely imaginary .

Commented by Tinkutara last updated on 10/Aug/17

Thank you very much Sir!