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If $z_1,z_2$ are two complex numbers such that $\large|\frac{z_1+z_2}{z_1-z_2}|=1,$ then $\large\frac{z_1}{z_2}$ is ?

$\begin{array}{1 1}(A) positive\; real \\(B) Negative \;real \\ (C) cannot\; say \\ (D) Zero\; or\; purely\; imaginary. \end{array}$

1 Answer

$\bigg|\large\frac{z_1+z_2}{z_1-z_2}\bigg|$$=1\Rightarrow\bigg|\large\frac{\large\frac{z_1}{z_2}+1}{\large\frac{z_1}{z_2}-1}\bigg|$$=1$
$=\bigg|\large\frac{x+iy+1}{x+iy-1}\bigg|$$=1$
$\Rightarrow\:\large\frac{\sqrt {(x+1)^2+y^2}}{\sqrt {(x-1)^2+y^2}}$$=1$
$\Rightarrow\:(x+1)^2+y^2=(x-1)^2+y^2$
$\Rightarrow\:(x+1)^2=(x-1)^2$
$\Rightarrow\:x=0,\:\:and\:\:y\in R$
$\Rightarrow\:\large\frac{z_1}{z_2}$ is either $0$ or is purely imaginary.
answered Jul 23, 2013 by rvidyagovindarajan_1
edited May 22, 2014 by rohanmaheshwari0831_1
 

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