Class 11Mathematics1 Complex Numbers and Quadratic Equations: Complex NumbersConjugate and Modulus of complex numbers
MediumIf $x+iy = \frac{a+ib}{a-ib}$,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 $x^2+y^2 = 1^2 = 1$.
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