If x + iy = sqrt{frac{a + ib}{c + id}},then ( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $x + iy = \sqrt{\frac{a + ib}{c + id}}$.Taking the conjugate on both sides,we get $x - iy = \sqrt{\frac{a - ib}{c - id}}$.Now,multiplying th

Vedclass · Accepted Answer

$\frac{a^2 + b^2}{c^2 + d^2}$

Vedclass · Answer

(A) Given $x + iy = \sqrt{\frac{a + ib}{c + id}}$.Taking the conjugate on both sides,we get $x - iy = \sqrt{\frac{a - ib}{c - id}}$.Now,multiplying these two equations:$(x + iy)(x - iy) = \sqrt{\frac{a + ib}{c + id}} 	imes \sqrt{\frac{a - ib}{c - id}}$$x^2 + y^2 = \sqrt{\frac{(a + ib)(a - ib)}{(c + id)(c - id)}}$$x^2 + y^2 = \sqrt{\frac{a^2 + b^2}{c^2 + d^2}}$Squaring both sides,we get $(x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2}$.

If $x + iy = \sqrt{\frac{a + ib}{c + id}}$,then $(x^2 + y^2)^2 = $

Similar Questions

If $\alpha_1, \alpha_2, \alpha_3$ respectively denote the moduli of the complex numbers $-i, \frac{1}{3}(1+i)$ and $-1+i$,then their increasing order is

The complex conjugate of $(4-3i)(2+3i)(1+4i)$ is

Let $z$ be a complex number such that $|z| + z = 3 + i$,where $i = \sqrt{-1}$. Then $|z| = $

The conjugate of the complex number $\frac{2 + 5i}{4 - 3i}$ is

If $x+iy=\sqrt{\frac{3+i}{1+3i}}$,then $\left(x^2+y^2\right)^2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module