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Question Number 32028 by rahul 19 last updated on 18/Mar/18

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

$${If}\:\frac{\mathrm{2}{z}_{\mathrm{1}} }{\mathrm{3}{z}_{\mathrm{2}} }\:{is}\:{a}\:{purely}\:{imaginary}\:{number}, \\ $$$${then}\:{find}\:{the}\:{value}\:{of}\:\mid\frac{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }{{z}_{\mathrm{1}} +{z}_{\mathrm{2}} }\mid\:. \\ $$

Answered by Giannibo last updated on 18/Mar/18

((2z_1 )/(3z_2 ))∈I ⇔ ((2z_1 )/(3z_2 ))=−((2z_1 ^(−) )/(3z_2 ^(−) ))⇔ z_1 z_2 ^(−) =−z_1 ^(−) z_2 (∗) ∣((z_1 −z_2 )/(z_1 +z_2 ))∣=w>0 ⇔ ∣((z_1 −z_2 )/(z_1 +z_2 ))∣^2 =w^2 ⇔ ((∣z_1 −z_2 ∣^2 )/(∣z_1 +z_2 ∣^2 ))=w^2 ⇔ (((z_1 −z_2 )(z_1 ^(−) −z_2 ^(−) ))/((z_1 +z_2 )(z_1 ^(−) +z_2 ^(−) )))=w^2 ⇔ ((z_1 z_1 ^(−) −z_2 z_1 ^(−) −z_1 z_2 ^(−) +z_2 z_2 ^(−) )/(z_1 z_1 ^(−) +z_2 z_1 ^(−) +z_1 z_2 ^(−) +z_2 z_2 ^(−) ))=w^2 ⇔^((∗)) ((z_1 z_1 ^(−) +z_2 z_2 ^(−) )/(z_1 z_1 ^(−) +z_2 z_2 ^(−) ))=w^2 ⇔ w^2 =1⇔ w=1

$$ \\ $$$$ \\ $$$$\frac{\mathrm{2}{z}_{\mathrm{1}} }{\mathrm{3}{z}_{\mathrm{2}} }\in\mathbb{I}\:\Leftrightarrow\:\frac{\mathrm{2}{z}_{\mathrm{1}} }{\mathrm{3}{z}_{\mathrm{2}} }=−\frac{\mathrm{2}\overline {{z}_{\mathrm{1}} }}{\mathrm{3}\overline {{z}_{\mathrm{2}} }}\Leftrightarrow\:{z}_{\mathrm{1}} \overline {{z}_{\mathrm{2}} }=−\overline {{z}_{\mathrm{1}} }{z}_{\mathrm{2}} \:\left(\ast\right) \\ $$$$ \\ $$$$\mid\frac{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }{{z}_{\mathrm{1}} +{z}_{\mathrm{2}} }\mid={w}>\mathrm{0}\:\Leftrightarrow \\ $$$$\mid\frac{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }{{z}_{\mathrm{1}} +{z}_{\mathrm{2}} }\mid^{\mathrm{2}} ={w}^{\mathrm{2}} \:\Leftrightarrow \\ $$$$\frac{\mid{z}_{\mathrm{1}} −{z}_{\mathrm{2}} \mid^{\mathrm{2}} }{\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid^{\mathrm{2}} }={w}^{\mathrm{2}} \:\Leftrightarrow \\ $$$$\frac{\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)\left(\overline {{z}_{\mathrm{1}} }−\overline {{z}_{\mathrm{2}} }\right)}{\left({z}_{\mathrm{1}} +{z}_{\mathrm{2}} \right)\left(\overline {{z}_{\mathrm{1}} }+\overline {{z}_{\mathrm{2}} }\right)}={w}^{\mathrm{2}} \:\Leftrightarrow \\ $$$$\frac{{z}_{\mathrm{1}} \overline {{z}_{\mathrm{1}} }−{z}_{\mathrm{2}} \overline {{z}_{\mathrm{1}} }−{z}_{\mathrm{1}} \overline {{z}_{\mathrm{2}} }+{z}_{\mathrm{2}} \overline {{z}_{\mathrm{2}} }}{{z}_{\mathrm{1}} \overline {{z}_{\mathrm{1}} }+{z}_{\mathrm{2}} \overline {{z}_{\mathrm{1}} }+{z}_{\mathrm{1}} \overline {{z}_{\mathrm{2}} }+{z}_{\mathrm{2}} \overline {{z}_{\mathrm{2}} }}={w}^{\mathrm{2}} \:\overset{\left(\ast\right)} {\Leftrightarrow} \\ $$$$\frac{{z}_{\mathrm{1}} \overline {{z}_{\mathrm{1}} }+{z}_{\mathrm{2}} \overline {{z}_{\mathrm{2}} }}{{z}_{\mathrm{1}} \overline {{z}_{\mathrm{1}} }+{z}_{\mathrm{2}} \overline {{z}_{\mathrm{2}} }}={w}^{\mathrm{2}} \:\Leftrightarrow \\ $$$${w}^{\mathrm{2}} =\mathrm{1}\Leftrightarrow \\ $$$${w}=\mathrm{1} \\ $$

Commented by rahul 19 last updated on 18/Mar/18

thank u sir!

$${thank}\:{u}\:{sir}! \\ $$

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