How to solve three complex equations

please find the attached file for an example of the system

Thanks in advance

[Torsten]

What does the symbol on the left-hand side of the equations mean ?

[Alex Sune]

@Torsten, this symbol indicates the phase of the complex number

[Walter Roberson]

Does the 1<0 (approximate rendering) indicate real value 1 complex phase 0, complex(1,0)?

Does j indicate sqrt(-1)?

Answer 1

[Alex Sune]

Easy using matrix notation:

A = [1.46j,0.9j,-0.36j;0.36,-0.42,-1.345;0.9j,1.65j,0.42];
b = [1;0;1];

I = A\b
I =

  -0.0059 - 0.4749i %I1
   0.0079 - 0.3480i %I2
  -0.0040 - 0.0184i %I3

[Najiya Omar]

Thanks Alex for your reply. your script dosent give the same anaswer in the textbook

I1=-j0.4839

I2=-j0.3357

I3 = -j0.0247

I was trying to use this script but still not giving the correct ansawer

syms I1 I2 I3
eqn1 = 1.46*I1 + 0.9*I2 - 0.36*I3 == 1;
eqn2 = 0.36*I1 - 0.42*I2 - 1.345*I3 == 0;
eqn3 = 0.9*I1 + 1.65*I2 + 0.42*I3 == 1;
sol = solve([eqn1, eqn2, eqn3], [I1, I2, I3]);
I1Sol = sol.I1
I2Sol = sol.I2
I3Sol = sol.I3

[Najiya Omar]

It is in fact giving same result, but in division formual not the last result. also it is not including -j

[Walter Roberson]

The left side of equations 1 and 3 need to be multiplied by 1i

[David Goodmanson]

Hi Najiya,

Alex's answer has a typo in that the last entry in A is 0.42 rather than j0.42. So:

A = [j*[1.46,0.9,-0.36];[0.36,-0.42,-1.345];j*[0.9,1.65,0.42]];
b = [1;0;1];
I = A\b
I =

   0.0000 - 0.4841i
   0.0000 - 0.3357i
   0.0000 - 0.0247i

I have no idea why I1 disagrees with what is in the book, although books have been known to be wrong. Unlike everything on this website.

[Najiya Omar]

Hi David,

You got it for me!! It worked now. Thank you very much. Greatly appreciated.