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

In the Argand plane,the quadrant in which $\frac{1+2i}{1-i}$ lies is:

Express the given complex number in the form $a+ib$: $(1-i)-(-1+i6)$

Let $A = \left\{ \theta \in \left( -\frac{\pi}{2}, \pi \right) : \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta} \text{ is purely imaginary} \right\}$. Then the sum of the elements in $A$ is

If $n$ is a positive integer and $\frac{(1+i)^n}{(1-i)^n} = -i$,then $n$ will be of the form:

Reduce $\left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{3-4 i}{5+i}\right)$ to the standard form.

Vedclass Test Series

Exam Paper Generator

Online Exam Module