Complex Number Problem

[varun_kanpur]

I am new to this forum.
Please help me with this problem-

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


On simplification, it comes

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

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

 

[mmm4444bot]

((1+i)x-2)/(3+i)+((2-3i)y+i)/(3-i)=i

On simplification, it [becomes]

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

Please check your typing. The equations above are not equivalent.

 

[varun_kanpur]

Please check your typing. The equations above are not equivalent.

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

 

[stapel]

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

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

I think something went wrong between these two lines. I don't get "-3i - 3" in the middle.

 

[Deleted member 4993]

If

A + B i = C + D i

then you can equate the coefficients of real parts to each other and equate the coefficients of imaginary parts to each other - to get -

A = C and B = D

 

[lookagain]

Please check your typing. The equations above are not equivalent.

That is a correct critique.

I think something went wrong between these two lines. I don't get "-3i - 3" in the middle.

And so is this critique above. I figured "-7 - 5i" to replace "-3i - 3."




So, then, in the style of equation you had, I got to:

4x -7 + 9y + i(2x - 5 - 7y) = 0


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So, applying what Subhotosh Khan posted, think of your equation as:


(4x + 9y - 7) + (2x - 7y - 5)i = 0 + 0i


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Then, your system of two equations in two unknowns becomes:


4x + 9y - 7 = 0

2x - 7y - 5 = 0 \(\displaystyle \ \ \ \ \ \ \ (This \ \ is \ \ after \ \ dividing \ \ both \ \ sides \ \ of \ \ the \ \ equation \ \) (2x - 7y - 5)i = 0i \(\displaystyle \ \ by \ \ i.)\)

 

[mmm4444bot]

Shouldn't that be 4x + 6y - 7 = 0?

The system that I worked out last night matches lookagain's.

The (x,y) solution checked out for me, in the original equation.

 

[varun_kanpur]

I thank everyone who helped me with this problem