If $x+iy = \frac{a+ib}{a-ib}$,prove that $x^{2}+y^{2}=1$.
(N/A) Given that $x+iy = \frac{a+ib}{a-ib}$.
Taking the modulus on both sides,we have $|x+iy| = \left|\frac{a+ib}{a-ib}\right|$.
Since $|z_1/z_2| = |z_1|/|z_2|$,we get $\sqrt{x^2+y^2} = \frac{|a+ib|}{|a-ib|}$.
We know that $|a+ib| = \sqrt{a^2+b^2}$ and $|a-ib| = \sqrt{a^2+(-b)^2} = \sqrt{a^2+b^2}$.
Thus,$\sqrt{x^2+y^2} = \frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}} = 1$.
Squaring both sides,we get $x^2+y^2 = 1^2 = 1$.
Taking the modulus on both sides,we have $|x+iy| = \left|\frac{a+ib}{a-ib}\right|$.
Since $|z_1/z_2| = |z_1|/|z_2|$,we get $\sqrt{x^2+y^2} = \frac{|a+ib|}{|a-ib|}$.
We know that $|a+ib| = \sqrt{a^2+b^2}$ and $|a-ib| = \sqrt{a^2+(-b)^2} = \sqrt{a^2+b^2}$.
Thus,$\sqrt{x^2+y^2} = \frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}} = 1$.
Squaring both sides,we get $x^2+y^2 = 1^2 = 1$.
Explore More
Similar Questions
If $x$ and $y$ are real numbers,which of the following is always true?
Medium
View SolutionLet complex number $z$ be such that $|z - \frac{6}{z}| = 5$,then the maximum value of $|z|$ will be -
${\left| {{z_1} + {z_2}} \right|^2} + {\left| {{z_1} - {z_2}} \right|^2}$ is equal to
DifficultAIEEE 2012
View SolutionFind the number of non-zero integral solutions of the equation $|1-i|^{x}=2^{x}$.
Medium
View SolutionThe modulus of the complex number $\frac{(1+i)^2(1+3 i)}{(2-6 i)(2-2 i)}$ is
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo