Values of x and y satisfying equation frac{{(1 + i)x - 2i}}{{3 + i}} + frac{{(2 - 3i),y + i}}{{3

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4-1.Complex numbers

easy

The values of $x$ and $y$ satisfying the equation $\frac{{(1 + i)x - 2i}}{{3 + i}}$ $ + \frac{{(2 - 3i)\,y + i}}{{3 - i}} = i$ are

$x = - 1,\,y = 3$

$x = 3,\,y = - 1$

$x = 0,\,y = 1$

$x = 1,y = 0$

(IIT-1980)

Solution

(b) $\frac{{(1 + i)x – 2i}}{{3 + i}} + \frac{{(2 – 3i)y + i}}{{3 – i}} = i$
==> $(4 + 2i)x + (9 – 7i)y – 3i – 3 = 10i$
Equating real and imaginary parts, we get $2x – 7y = 13$ and $4x + 9y = 3$. Hence $x = 3$and $y = – 1$.
Trick : After finding the equations, no need to solve them, put the values of $x$ and $y$ given in the options and get the appropriate option.

Standard 11

Mathematics

If $\,\left| \begin{array}{l}\,6i\,\,\,\,\, – 3i\,\,\,\,\,\,\,\,\,1\\\,\,4\,\,\,\,\,\,\,\,\,3i\,\,\,\,\,\, – 1\\\,20\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,i\end{array} \right|\,$=$x + iy$, then $(x, y)$ is

easy

If the real part of the complex number $(1-\cos \theta+2 i \sin \theta)^{-1}$ is $\frac{1}{5}$ for $\theta \in(0, \pi)$, then the value of the integral $\int_{0}^{\theta} \sin x \,d x$ is equal to:

hard

(JEE MAIN-2021)

If $\frac{{5( – 8 + 6i)}}{{{{(1 + i)}^2}}} = a + ib$, then$(a,\,b)$ equals

easy

The real values of $x$ and $y$for which the equation is $(x + iy)$ $(2 – 3i)= 4 + i$ is satisfied, are

easy

The number of real values of $a$ satisfying the equation ${a^2} – 2a\sin x + 1 = 0$ is

medium

The values of $x$ and $y$ satisfying the equation $\frac{{(1 + i)x - 2i}}{{3 + i}}$ $ + \frac{{(2 - 3i)\,y + i}}{{3 - i}} = i$ are

Solution

Similar Questions

If $\,\left| \begin{array}{l}\,6i\,\,\,\,\, – 3i\,\,\,\,\,\,\,\,\,1\\\,\,4\,\,\,\,\,\,\,\,\,3i\,\,\,\,\,\, – 1\\\,20\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,i\end{array} \right|\,$=$x + iy$, then $(x, y)$ is

If the real part of the complex number $(1-\cos \theta+2 i \sin \theta)^{-1}$ is $\frac{1}{5}$ for $\theta \in(0, \pi)$, then the value of the integral $\int_{0}^{\theta} \sin x \,d x$ is equal to:

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