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Question Number 140443 by EnterUsername last updated on 07/May/21

If z_{2} / z_{1} is purely imaginary and a and b are non - zero real

numbers, then ∣ ({a z}_{1} + {b z}_{2}) / ({a z}_{1} - {b z}_{2}) ∣ is equal to_____.

Answered by mr W last updated on 07/May/21

z_{2} = r_{2} e^{θ_{2} i}

z_{1} = r_{1} e^{θ_{1} i}

\frac{z_{2}}{z_{1}} = (\frac{r_{2}}{r_{1}}) e^{(θ_{2} - θ_{1}) i} = i m a g i n a r y

\Rightarrow θ_{2} - θ_{1} = \frac{(2 k + 1) π}{2}

\frac{{a z}_{1} + {b z}_{2}}{{a z}_{1} - {b z}_{2}} = \frac{{a r}_{1} e^{θ_{1} i} + {b r}_{2} e^{θ_{2} i}}{{a r}_{1} e^{θ_{1} i} - {b r}_{2} e^{θ_{2} i}}

= \frac{{a r}_{1} + {b r}_{2} e^{(θ_{2} - θ_{1}) i}}{{a r}_{1} - {b r}_{2} e^{(θ_{2} - θ_{1}) i}}

= \frac{{a r}_{1} + {b r}_{2} e^{\frac{2 k + 1}{2} i}}{{a r}_{1} - {b r}_{2} e^{\frac{2 k + 1}{2} i}}

= \frac{{a r}_{1} \pm {b r}_{2} i}{{a r}_{1} \mp {b r}_{2} i}

= \frac{z_{3}}{{\bar{z}}_{3}}

= \frac{r_{3} e^{θ_{3} i}}{r_{3} e^{- θ_{3} i}}

= e^{2 θ_{3} i}

∣ \frac{{a z}_{1} + {b z}_{2}}{{a z}_{1} - {b z}_{2}} ∣=∣ e^{2 θ_{3} i} ∣= 1

Commented by EnterUsername last updated on 07/May/21

Thank you, Sir