If z_2 /z_1 is purely imaginary and a and b are non-zero real numbers, then ∣(az_1 +bz_2 )/(az_1 −bz_2 )∣ is equal to ____


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
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

	$Thank you, Sir$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum