Complex Numbers

1. Jun 6, 2006

I'm not sure if this even belongs in the calculus forum because it is rather basic, but here it goes.

I'm working on this circuit problem and I could NOT figure out what I was doing wrong.

My solution has led me to the following system of equations:

(1) $$(2+j2)\bar I_1 +(j1)\bar I_2 = 10$$
(2) $$(j)\bar I_1 +(2+j2)\bar I_2 = 0$$

I then bought a cramster account since they have the solutions for my book. Their solution is the same system of equations (1) and (2).

HOWEVER. When the book AND the cramster solution solves for $\bar I_2$ they yield the following:
(by the way I just need $\bar I_2$ to solve the problem)

$$\bar I_2 = \frac{10}{-2-j} =-4+2j$$

Now when I solve the following matrix (with the help of a calculator) I have:
$$\left( \begin{array}{cc} 2+j2 & j \\ j & 2+j2 \end{array} \right) \left( \begin{array}{c} \bar I_1 \\ \bar I_2 \end{array} \left) = \left( \begin{array}{c} 10 \\ 0 \end{array} \right)$$

Solving yields the following result:
(1*) $$\bar I_1 = \frac{36}{13} - \frac{28}{13}j$$
(2*) $$\bar I_2 = -\frac{16}{13} - \frac{2}{13}j$$

Now if I plug (1*) and (2*) back into (1) and (2) I get $10$ and $0$ respectively.

So what gives? Is the book and cramster wrong?
I haven't taken complex analysis, so maybe there are more then one solutions to $\bar I_2$ How do I know which one is right?

grr... I was tearing my hair out thinking I did something wrong.

Last edited: Jun 6, 2006
2. Jun 6, 2006

TD

If you want to compare your solution with the one you found, they have to be in the same form. The cramster solution is not yet in the form a+bj, you need to multiply numerator and denominator with the complement of the denominator. This should give -4+2j, still not the same as your answer - although I see nothing wrong with your system, matrix and solution.

3. Jun 6, 2006

My bad on that. I should have shown it in comparable forms (I did this, but didn't post that work on here). I'll edit my post to reflect the change.

Let me refine my question a bit.

Given the system of equations
$$(2+j2)\bar I_1 +(j1)\bar I_2 = 10$$
$$(j)\bar I_1 +(2+j2)\bar I_2 = 0$$

Is it possible for $\bar I_2 = -4+2j$ to be a solution?

For example, if I was given the following system of equations:
$$y = x$$
$$y = -x +10$$

I know that the only solutions are: $$x=5$$ and $$x=-5$$. Since I've already shown that (1*) and (2*) [from my original post] are solutions, can I simply conclude that $\bar I_2 = -4+2j$ is incorrect.

I just can't imagine the book and an independent website both being wrong, and I'm also interested to know how solutions form with complex numbers.

Last edited: Jun 6, 2006
4. Jun 6, 2006

shmoe

Your coefficient matrix is invertible-> there is a unique solution.

You can check their answer by substituting it into both equations and solving for $$\bar I_1$$, looks like you get different answers.

Maybe cramster just used what the book had.

5. Jun 6, 2006