If x, y in R and (x + iy)(3 + 2i) = 1 + i,the | vedclass

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

Question

(C) Given $(x + iy)(3 + 2i) = 1 + i$. $x + iy = \frac{1 + i}{3 + 2i}$. To simplify,multiply the numerator and denominator by the conjugate of the deno

Vedclass · Accepted Answer

$\left( \frac{5}{13}, \frac{1}{13} \right)$

Vedclass · Answer

(C) Given $(x + iy)(3 + 2i) = 1 + i$. $x + iy = \frac{1 + i}{3 + 2i}$. To simplify,multiply the numerator and denominator by the conjugate of the denominator,$(3 - 2i)$: $x + iy = \frac{(1 + i)(3 - 2i)}{(3 + 2i)(3 - 2i)} = \frac{3 - 2i + 3i - 2i^2}{3^2 + 2^2}$. Since $i^2 = -1$,we have: $x + iy = \frac{3 + i - 2(-1)}{9 + 4} = \frac{3 + i + 2}{13} = \frac{5 + i}{13} = \frac{5}{13} + i\frac{1}{13}$. Comparing the real and imaginary parts,we get $x = \frac{5}{13}$ and $y = \frac{1}{13}$. Thus,$(x, y) = \left( \frac{5}{13}, \frac{1}{13} \right)$.

If $x, y \in R$ and $(x + iy)(3 + 2i) = 1 + i$,then $(x, y)$ is

Similar Questions

If $\left(\frac{1+i}{1-i}\right)^m=1$,then $m$ cannot be equal to

$A$ value of $\theta$ for which $\frac{2 + 3i \sin \theta}{1 - 2i \sin \theta}$ is purely imaginary,is:

The value of $(1 + i)^8 + (1 - i)^8$ is

If $(2x - y + 1) + i(x - 2y - 1) = 2 - 3i$,then the multiplicative inverse of $(x - iy)$ is

Express the given complex number in the form $a+ib$: $i^{9}+i^{19}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module