Time for BG help on some math regarding the complex number

[Osede]

This thing just driving me mad. Let say that we have these two equations

(875 + j300) X - (j1050) Y = 7580 + j 4375
-50 X + (50-25j) Y = 0


And both X and Y consist of real and complex number. Suppose the proper way to do it is
expand the X and Y into

X = real_x + imag_x * j
Y = real_y + imag_y * j

And substitute them into the equation, then solve by seperate the real and imaginary part.

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Ok, short story is, the solution I got from manual is

X = 3.7866 + j4.9598
Y = 1.0456 + j5.4812

While using inverse matrix method with matlab giving this solution...

X = 0.2746 + j0.8917
Y = -3.3469 + j7.243


Theoretically both solution seems to be correct (as approximation). But I just want to ask if anyone know the other way that will lead me to obtain the same solution as solution manual? (the first set of solution)

Since this work is to confirm and making sure that the answer from solution manual is correct before giving away to student. However, even thought I know who are incharge on this problem, but it is impossible to get to him at any moment

[thestalkmore]

42

[Blubbartron]

Are you using j to repesent the square root of -1, also known as i? Your notation is confusing the shit out of me. A number in the set Z (set of complex numbers) is typically expressed as a + bi. I'm also at a loss as to why you used matrices to solve a simple substitution method two equations two unknowns problem.

edit: On another note, solving two equations two unknowns doesn't produce "approximations." The numbers you find either work, or they don't. Then again, I can't check either of the answers provided as you haven't given us the real values of x and y.

[Creeps]

Every day, I wake up and thank god(s) that I'm not a mathematician

[Korietsu]

If you're using the EE way of solving the equations j stands for i = √-1 Then you can just use DeMoivre's Thereom to solve iirc.

[Blubbartron]

Oh, that sounds probable. My linear algebra class was extremely sub-par.

[Osede]

Are you using j to repesent the square root of -1, also known as i? Your notation is confusing the shit out of me. A number in the set Z (set of complex numbers) is typically expressed as a + bi. I'm also at a loss as to why you used matrices to solve a simple substitution method two equations two unknowns problem.

yea, j = i = square root of -1. It just comes across that this book is using j as complex number.

edit: On another note, solving two equations two unknowns doesn't produce "approximations." The numbers you find either work, or they don't. Then again, I can't check either of the answers provided as you haven't given us the real values of x and y.

the reason I was saying it is approximation is that there is still like 0.1~0.5% of error from those solutions >.>

If you're using the EE way of solving the equations j stands for i = √-1 Then you can just use DeMoivre's Thereom to solve iirc.

Thanks, I will look into this one

[The_OG_Nelta]

inb4 aurik

[Radec]

Confirming the book on this one.

-50 X + (50-25j) Y = 0
X = (50-25j)*Y/50

(875 + 300j)*(50-25j)*Y/50 - (1050j) Y = 7580 + 4375j
(((875+300j)*(1-0.5j) - 1050j)*Y) = 7580 + 4375j
(875-(875/2)j+300j-(300/2)(j^2)-1050j)*Y = 7580 + 4375j
(1025-1187.5j)*Y = 7580 + 4375j

At this point if I had to do it by hand I'd split it to real/imaginary parts but since you have matlab, just paste this in to confirm

(7580 + 4375*sqrt(-1))/(1025-1187.5*sqrt(-1))

ans = 1.0461 + 5.4802i = Y

Substitute back in and

(50-25j)*(1.0461 + 5.4802j)/50

ans = 3.7862 + 4.9571i = X

...which doesn't match the book exactly, but much closer than the other solution - if I put your alternative numbers into the left sides of the original equations, I get

(1) = (7.5779e+003 +4.3769e+003i) ok, this is close, but...
(2) = (1.7764e-014 +4.0124e+002i) instead of 0

[Demosthenes11]

thats just like solving a system of solutions
aka
3x-2y=2
x+5y=7

or something like that (that is, if im reading this correctly)