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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-

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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 ∣(az_1 +bz_2 )/(az_1 −bz_2 )∣ is equal to _____.
$$\mathrm{If}\:{z}_{\mathrm{2}} /{z}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{purely}\:\mathrm{imaginary}\:\mathrm{and}\:{a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{non}-\mathrm{zero}\:\mathrm{real} \\ $$$$\mathrm{numbers},\:\mathrm{then}\:\mid\left({az}_{\mathrm{1}} +{bz}_{\mathrm{2}} \right)/\left({az}_{\mathrm{1}} −{bz}_{\mathrm{2}} \right)\mid\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\_\_\_\_\_. \\ $$
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) (z_2 /z_1 )=((r_2 /r_1 ))e^((θ_2 −θ_1 )i) =imaginary ⇒θ_2 −θ_1 =(((2k+1)π)/2) ((az_1 +bz_2 )/(az_1 −bz_2 ))=((ar_1 e^(θ_1 i) +br_2 e^(θ_2 i) )/(ar_1 e^(θ_1 i) −br_2 e^(θ_2 i) )) =((ar_1 +br_2 e^((θ_2 −θ_1 )i) )/(ar_1 −br_2 e^((θ_2 −θ_1 )i) )) =((ar_1 +br_2 e^(((2k+1)/2)i) )/(ar_1 −br_2 e^(((2k+1)/2)i) )) =((ar_1 ±br_2 i)/(ar_1 ∓br_2 i)) =(z_3 /z_3 ^� ) =((r_3 e^(θ_3 i) )/(r_3 e^(−θ_3 i) )) =e^(2θ_3 i) ∣((az_1 +bz_2 )/(az_1 −bz_2 ))∣=∣e^(2θ_3 i) ∣=1
$${z}_{\mathrm{2}} ={r}_{\mathrm{2}} {e}^{\theta_{\mathrm{2}} {i}} \\ $$$${z}_{\mathrm{1}} ={r}_{\mathrm{1}} {e}^{\theta_{\mathrm{1}} {i}} \\ $$$$\frac{{z}_{\mathrm{2}} }{{z}_{\mathrm{1}} }=\left(\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }\right){e}^{\left(\theta_{\mathrm{2}} −\theta_{\mathrm{1}} \right){i}} ={imaginary} \\ $$$$\Rightarrow\theta_{\mathrm{2}} −\theta_{\mathrm{1}} =\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}} \\ $$$$\frac{{az}_{\mathrm{1}} +{bz}_{\mathrm{2}} }{{az}_{\mathrm{1}} −{bz}_{\mathrm{2}} }=\frac{{ar}_{\mathrm{1}} {e}^{\theta_{\mathrm{1}} {i}} +{br}_{\mathrm{2}} {e}^{\theta_{\mathrm{2}} {i}} }{{ar}_{\mathrm{1}} {e}^{\theta_{\mathrm{1}} {i}} −{br}_{\mathrm{2}} {e}^{\theta_{\mathrm{2}} {i}} } \\ $$$$=\frac{{ar}_{\mathrm{1}} +{br}_{\mathrm{2}} {e}^{\left(\theta_{\mathrm{2}} −\theta_{\mathrm{1}} \right){i}} }{{ar}_{\mathrm{1}} −{br}_{\mathrm{2}} {e}^{\left(\theta_{\mathrm{2}} −\theta_{\mathrm{1}} \right){i}} } \\ $$$$=\frac{{ar}_{\mathrm{1}} +{br}_{\mathrm{2}} {e}^{\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}}{i}} }{{ar}_{\mathrm{1}} −{br}_{\mathrm{2}} {e}^{\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}}{i}} } \\ $$$$=\frac{{ar}_{\mathrm{1}} \pm{br}_{\mathrm{2}} {i}}{{ar}_{\mathrm{1}} \mp{br}_{\mathrm{2}} {i}} \\ $$$$=\frac{{z}_{\mathrm{3}} }{\bar {{z}}_{\mathrm{3}} } \\ $$$$=\frac{{r}_{\mathrm{3}} {e}^{\theta_{\mathrm{3}} {i}} }{{r}_{\mathrm{3}} {e}^{−\theta_{\mathrm{3}} {i}} } \\ $$$$={e}^{\mathrm{2}\theta_{\mathrm{3}} {i}} \\ $$$$\mid\frac{{az}_{\mathrm{1}} +{bz}_{\mathrm{2}} }{{az}_{\mathrm{1}} −{bz}_{\mathrm{2}} }\mid=\mid{e}^{\mathrm{2}\theta_{\mathrm{3}} {i}} \mid=\mathrm{1} \\ $$
Commented by EnterUsername last updated on 07/May/21
Thank you, Sir
$$\mathrm{Thank}\:\mathrm{you},\:\mathrm{Sir} \\ $$

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