If ((2z_1 )/(3z_2 )) is a purely imaginary number, then find the value of ∣((z_1 −z_2 )/(z_1 +z


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 32028 by rahul 19 last updated on 18/Mar/18

$I f \frac{2 z_{1}}{3 z_{2}} i s a p u r e l y i m a g i n a r y n u m b e r,$ $t h e n f i n d t h e v a l u e o f ∣ \frac{z_{1} - z_{2}}{z_{1} + z_{2}} ∣ .$
	Answered by Giannibo last updated on 18/Mar/18

	$\frac{2 z_{1}}{3 z_{2}} \in I \Leftrightarrow \frac{2 z_{1}}{3 z_{2}} = - \frac{2 \overset{―}{z_{1}}}{3 \overset{―}{z_{2}}} \Leftrightarrow z_{1} \overset{―}{z_{2}} = - \overset{―}{z_{1}} z_{2} ()$ $∣ \frac{z_{1} - z_{2}}{z_{1} + z_{2}} ∣= w > 0 \Leftrightarrow$ $∣ \frac{z_{1} - z_{2}}{z_{1} + z_{2}} ∣^{2} = w^{2} \Leftrightarrow$ $\frac{∣ z_{1} - z_{2} ∣^{2}}{∣ z_{1} + z_{2} ∣^{2}} = w^{2} \Leftrightarrow$ $\frac{(z_{1} - z_{2}) (\overset{―}{z_{1}} - \overset{―}{z_{2}})}{(z_{1} + z_{2}) (\overset{―}{z_{1}} + \overset{―}{z_{2}})} = w^{2} \Leftrightarrow$ $\frac{z_{1} \overset{―}{z_{1}} - z_{2} \overset{―}{z_{1}} - z_{1} \overset{―}{z_{2}} + z_{2} \overset{―}{z_{2}}}{z_{1} \overset{―}{z_{1}} + z_{2} \overset{―}{z_{1}} + z_{1} \overset{―}{z_{2}} + z_{2} \overset{―}{z_{2}}} = w^{2} \overset{()}{\Leftrightarrow}$ $\frac{z_{1} \overset{―}{z_{1}} + z_{2} \overset{―}{z_{2}}}{z_{1} \overset{―}{z_{1}} + z_{2} \overset{―}{z_{2}}} = w^{2} \Leftrightarrow$ $w^{2} = 1 \Leftrightarrow$ $w = 1$
	Commented by rahul 19 last updated on 18/Mar/18

	$t h a n k u s i r!$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum