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If 2z1/3z2 is purely imaginary then , | Z 1 – z 2 / z 1 + z 2 | = a) 3/2 b) 1 c) 2/3 d) 4/9


If 2z1/3z2  is  purely  imaginary  then, | Z1 – z2/ z1 + z2| =


a)      3/2


b)      1


c)      2/3


d)      4/9


Grade:12

2 Answers

Aman Bansal
592 Points
12 years ago

Dear Pooja,

\text{Say } z_1 \text{ is }  r_1e^{i\theta_1} \text{ and }z_2 \text{ is }  r_2e^{i\theta_2} \text{ so }\\\\ \frac{2z_1}{3z_2}=\frac{2r_1}{3r_2}e^{i(\theta_1 - \theta_2)}\\\\ \text{Now if that is imaginary then } \theta_1 - \theta_2  = (2n+1)\frac{\pi}{2}\\\\ \text{Meaning they both are perpendicular to each other so the mod of }\\\\ z_1+z_2 \text{ and } z_1-z_2 \text{ will be }\\\\ \sqrt{{|z_1|}^2+{|z_2|}^2}\\\\ \text{Hence the ans is 1}

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Ashwin Muralidharan IIT Madras
290 Points
12 years ago

Hi Pooja,

 

If 2z1/3z2 is purely imaginary, then z1/z2 is purely imaginary.

Which means z1/z2 + z1'/z2' = 0 -------------(where ' denotes the complex conjugate)

So z1z2' + z1'z2 = 0--------------------(1)

 

Now | (z1-z2)/(z1+z2) |^2 = (z1-z2)(z1'-z2')/(z1+z2)(z1'+z2') = {|z1|^2 + |z2|^2}/{|z1|^2 + |z2|^2} = 1 --------[By using (1)].

So the given expression = 1. Option (2).

 

Best Regards,

Ashwin (IIT Madras).

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