1. ## Complex Number Problem

I am new to this forum.

Find the real values of x and y
if
((1+i)x-2)/(3+i)+((2-3i)y+i)/(3-i)=i where i = imaginary number

2. ## Re: Complex Number Problem

The first step is to simplify the fraction so that there is no imaginary part below the line. To do this multiply the top and bottom of the fraction by the complex conjugate of the bottom.

3. ## Re: Complex Number Problem

Originally Posted by Shakarri
The first step is to simplify the fraction so that there is no imaginary part below the line. To do this multiply the top and bottom of the fraction by the complex conjugate of the bottom.
On simplification, it comes

4x + 9y + i(2x-13-7)=0

What to do for getting real values of x and y ?

4. ## Re: Complex Number Problem

What I did is-

$\displaystyle \frac{(1+i)x-2}{3+i}+\frac{(2-3i)y+i}{3-i}= i$

$\displaystyle \frac{4x +2xi-3i-3+9y-7yi}{10}=i$

$\displaystyle 4x +2xi-3i-3+9y-7yi=10i$

$\displaystyle 4x +9y+i(2x-13-7y)=0$

What to do to find real values of x and y

5. ## Re: Complex Number Problem

Originally Posted by varunkanpur
What I did is-
$\displaystyle \frac{(1+i)x-2}{3+i}+\frac{(2-3i)y+i}{3-i}= i$
$\displaystyle 4x +9y+i(2x-13-7y)=0$
What to do to find real values of x and y
Solve this system:
$\displaystyle \begin{cases}4x +9y=0\\2x-13-7y=0\end{cases}$