If (1 - i)x + (1 + i)y = 1 - 3i,then (x, y) = | vedclass

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

Question

(A) Given the equation: $(1 - i)x + (1 + i)y = 1 - 3i$Expanding the terms,we get: $(x - ix) + (y + iy) = 1 - 3i$Grouping real and imaginary parts: $(x

Vedclass · Accepted Answer

$(2, -1)$

Vedclass · Answer

(A) Given the equation: $(1 - i)x + (1 + i)y = 1 - 3i$Expanding the terms,we get: $(x - ix) + (y + iy) = 1 - 3i$Grouping real and imaginary parts: $(x + y) + i(y - x) = 1 - 3i$Equating the real and imaginary parts on both sides:$x + y = 1$  (Equation $1$)$y - x = -3$ (Equation $2$)Adding Equation $1$ and Equation $2$:$(x + y) + (y - x) = 1 + (-3)$$2y = -2$$y = -1$Substituting $y = -1$ into Equation $1$:$x + (-1) = 1$$x = 2$Therefore,$(x, y) = (2, -1)$.

If $(1 - i)x + (1 + i)y = 1 - 3i$,then $(x, y) = $

Similar Questions

The solution of the equation $|z| - z = 1 + 2i$ is

If $(x+iy)^{\frac{1}{3}} = 5+3i$,then $3x+5y = $

If $|a| = a$ and $|b| = b$,then $\left( \frac{a}{a^2} - \frac{b}{b^2} \right)^2 = $

$\frac{(1+i)^{2011}}{(1-i)^{2009}}$ is equal to

$\left(\frac{1+i}{1-i}\right)^4+\left(\frac{1-i}{1+i}\right)^4$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module