Solving systems of equations that contain complex numbers

In summary, The conversation discusses solving systems of equations containing complex numbers in the context of circuit theory and phasors. The goal is to find I2 and Voc, and the equations can be converted to vector-matrix form for easier solving. The standard methods of Gaussian elimination, matrix inversion, or row-reduction can be used, but with complex arithmetic instead of real arithmetic. It is noted that the given answers may be incorrect and alternative methods can be used.
  • #1
Cocoleia
295
4

Homework Statement


I am having trouble solving systems of equations when they contain complex numbers. The context is circuit theory and phasors. For example, I am given this
upload_2016-12-18_10-7-43.png

And the goal is to find I2 and Voc, which you can see the answers for. I just don't know how to manipulate the numbers to get to this answer. Can someone explain the steps given these equations?
 
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  • #2
You can convert the values with angles to complex values with real and imaginary part as well, and then solve in the same way you would do it with real numbers.
 
  • #3
Cocoleia said:

Homework Statement


I am having trouble solving systems of equations when they contain complex numbers. The context is circuit theory and phasors. For example, I am given this
View attachment 110531
And the goal is to find I2 and Voc, which you can see the answers for. I just don't know how to manipulate the numbers to get to this answer. Can someone explain the steps given these equations?

In vector-matrix form, your equations read as
$$\pmatrix{600-300j&300j&0\\-300j&300+300j&-2\\300j& -300j&1} \pmatrix{I_1\\I_2\\V} = \pmatrix{9\\0\\0}$$.
This is just an ordinary 3x3 linear system, that you can solve using Gaussian elimination or matrix inversion or row-reduction---all standard elementary algebra methods. (The only difference is that you need to use complex arithmetic instead of real arithmetic.)

When I solve this system using Maple I get a solution much different from the one you propose.
 
  • #4
Ray Vickson said:
In vector-matrix form, your equations read as
$$\pmatrix{600-300j&300j&0\\-300j&300+300j&-2\\300j& -300j&1} \pmatrix{I_1\\I_2\\V} = \pmatrix{9\\0\\0}$$.
This is just an ordinary 3x3 linear system, that you can solve using Gaussian elimination or matrix inversion or row-reduction---all standard elementary algebra methods. (The only difference is that you need to use complex arithmetic instead of real arithmetic.)

When I solve this system using Maple I get a solution much different from the one you propose.
These were the answers given to us by the professor. They could be wrong. But thanks, I will try to solve it like that
 

FAQ: Solving systems of equations that contain complex numbers

1. How do I solve a system of equations containing complex numbers?

To solve a system of equations containing complex numbers, you will use the same methods as solving a system of equations with real numbers. You will need to isolate the variable in one equation and then substitute that value into the other equation to solve for the other variable.

2. Can I use the quadratic formula to solve systems of equations with complex numbers?

Yes, you can use the quadratic formula to solve systems of equations with complex numbers. The only difference is that you will be working with complex numbers instead of real numbers. You will still have to take the square root of a complex number, which will result in two solutions (one positive and one negative).

3. Do I need to use the imaginary unit "i" when solving systems of equations with complex numbers?

Yes, you will need to use the imaginary unit "i" when solving systems of equations with complex numbers. This is because complex numbers are in the form a + bi, where "a" is the real part and "bi" is the imaginary part. "i" represents the square root of -1 and is essential in working with complex numbers.

4. What is the difference between a real solution and a complex solution?

A real solution is a solution that is a real number, meaning it does not involve the imaginary unit "i". A complex solution is a solution that involves the imaginary unit "i". In other words, a complex solution is a combination of both real and imaginary numbers.

5. Can a system of equations with complex numbers have more than two solutions?

Yes, a system of equations with complex numbers can have more than two solutions. This is because complex numbers have two parts (real and imaginary), so there can be multiple combinations of values that satisfy the equations. However, a system of two equations can only have a maximum of two unique solutions.

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