# Algebra with complex

1. Sep 9, 2014

### ChasW.

1. The problem statement, all variables and given/known data

I am following along in a book which provides the equation below.
I have also included the book's stated solution.

2. Relevant equations

The Equation:
$${R_T+jX_T \over 1+jB_C(R_T+jX_T)}+jX_L = R_A-jX_A$$
The Book's Solution:
$$R_T=R_A(1-B_CX_T)+(X_A+X_L)B_CR_T$$
3. The attempt at a solution

When attempting to solve for RT, so far I have:
$${R_T+jX_T \over 1+jB_CR_T-B_CX_T} = R_A-jX_A-jX_L$$
$$R_T=(R_A-jX_A-jX_L)(1+jB_CR_T-B_CX_T)-jX_T$$
$$R_T= R_A+R_AjB_CR_T-R_AB_CX_T -jX_A+X_AB_CR_T+jX_AB_CX_T -jX_L+X_LB_CR_T+jX_LB_CX_T -jX_T$$
I then begin to simplify:
$$R_T= R_A(1+jB_CR_T-B_CX_T) +X_A(-j+B_CR_T+jB_CX_T) +X_L(-j+B_CR_T+jB_CX_T) -jX_T$$
$$R_T= R_A(1+jB_CR_T-B_CX_T) +(X_A+X_L)(-j+B_CR_T+jB_CX_T) -jX_T$$
I seem to be having some issue identifying further simplification.
Any assistance would be appreciated.
Charles

2. Sep 9, 2014

### Staff: Mentor

Could you give more context? It appears that at some point only the real part was retained.

3. Sep 9, 2014

### ChasW.

You are exactly correct. The greater context is a series of equations describing an impedance match using a series inductor and parallel capacitor from RF Circuit Design Theory and Applications where RT is the real part. The example starts off as expressing impedances as real and imaginary components and then separating the two: RT and XT.

The book's stated XT solution is:
$$X_T=R_TR_AB_C-(1-B_CX_T)(X_A+X_L)$$
Perhaps my confusion resides with solving for the real-only part. If all that was done was discarding the imaginary parts in the RT equation, then I might be ok.

My reason for wanting to clearly understand the equation and its transformations is to be able to perform the same approach but as applied to other topologies such as series capacitor and parallel inductor.

The way the equations RT and XT are used is described as:
In the RT equation, solve for XL and substitute those results into the XT equation, solve the remaining quadratic equation (plus side version) to obtain BC and then substitute BC into linear XT equation to solve for XL.

Last edited: Sep 9, 2014
4. Sep 9, 2014

### nasu

You need to equate the real and imaginary parts of the complex equation separately.
I mean the real left-hand side equals the real right-hand side and same for the imaginary (if needed).
You can do some simplifications first, as you did. But in the end equate the two parts separately.

5. Sep 9, 2014

### ChasW.

Ok so after the simplifications that I did, does equating to the real part simply mean to drop all of the remaining imaginary components to arrive at the form of RT that the book author did? Or perhaps to reword my question, does taking the real part of RT at the point where I simplified to, mean discarding all remaining imaginary values? I guess what I was expecting was some mathematical step beyond my final simplification or more methodical means for taking the real part of the lengthy expression.

6. Sep 9, 2014

### nasu

It's not really "discarding". You just use the general meaning of two complex numbers being equal.
If you have z1=a1+b1*i and z2=a2+b2*i (where a1,a2,b1,b2 are pure real), the equality z1=z2 implies
a1=a2
and
b1=b2.

Here z1 and z2 are the two sides of your equation.

7. Sep 9, 2014

### ChasW.

Ok I get the basic example you've given as it demonstrates the equality of the real and imaginary parts of two different complex numbers given z1=z2. I am still working towards properly applying that to the separation of the real and imaginary components in the larger expressions.

So for separating out the imaginary side which the book states to be:
$$X_T=R_TR_AB_C-(1-B_CX_T)(X_A+X_L)$$
which comes from:
$${R_T+jX_T \over 1+jB_C(R_T+jX_T)}+jX_L = R_A-jX_A$$
I can take it as far as:
$${R_T+jX_T \over 1+jB_CR_T-B_CX_T} = R_A-jX_A-jX_L$$
$$jX_T=(R_A-jX_A-jX_L)(1+jB_CR_T-B_CX_T)-R_T$$
$$jX_T=R_A+R_AjB_CR_T-R_AB_CX_T -jX_A+X_AB_CR_T+jX_AB_CX_T -jX_L+X_LB_CR_T+jX_LB_CX_T-R_T$$
$$jX_T=R_A(1+jB_CR_T-B_CX_T)+X_A(-j+B_CR_T+jB_CX_T)+X_L(-j+B_CR_T+jB_CX_T)-R_T$$
$$jX_T=R_A(1+jB_CR_T-B_CX_T)+(X_A+X_L)(-j+B_CR_T+jB_CX_T)-R_T$$
What is the next step towards separating out XT?

8. Sep 9, 2014

### nasu

You need to work more on the right side.
Collect all the terms with j. This will be the imaginary part of the left hand side. (Im)
Collect all terms without j. This will be the real part. (Re)

The way you wrote it so far, you have only imaginary part on the left side.
So after you do the above, you will have
XT=Im
and
0=Re.

Solve the two equations for whatever you want to find.

9. Sep 9, 2014

### ChasW.

When I solve for XT I get:
$$X_T={R_T+jX_T \over R_A(-jB_C^2R_T)+(X_A+X_L)(jB_C-B_C^2R_T)}$$
which doesn't seem to be much better than the last attempt.
Perhaps you would be willing to show me the transformation to XT=Im as you have described?

10. Sep 9, 2014

### Staff: Mentor

You're mixing real and imaginary again.

Consider
\begin{align} \frac{A + j B}{C + j D} &= F + jG \\ A + j B &= (C + j D)(F + jG) \\ A + j B &= CF + j CG + j DF - DG \\ A + j B &= (CF - DG) + j (CG + DF) \end{align}
Look at that last line. It means that $A = CF - DG$ and $B = CG + DF$. You need to do something similar with your equation.

11. Sep 9, 2014

### ChasW.

I think this helped.

Starting from
$$R_T+jX_T=R_A(1+jB_CR_T-B_CX_T)+(X_A+X_L)(-j+B_CR_T+jB_CX_T)$$
Using what you've provided I can see that
$$R_T=R_A(1-B_CX_T)+(X_A+X_L)(B_CR_T)$$
which also brings me closer with XT
$$X_T=R_AB_CR_T+(X_A+X_L)(-1+B_CX_T)$$
If I can just find the sign error now. For XT, the book has
$$X_T=R_TR_AB_C-(1-B_CX_T)(X_A+X_L)$$
... or is that actually not an error, but rather adding the opposite? Mental fatigue has bested me :)

Thank you both.
Charles

Last edited: Sep 9, 2014
12. Sep 9, 2014

### Staff: Mentor

When your brain has rested, I'm sure you'll see that
$$(-1+B_CX_T) (X_A+X_L) = -(1-B_CX_T) (X_A+X_L)$$

13. Sep 9, 2014

### nasu

It's not an error. They are the same.
Take the "-" outside the parenthesis and you get the result in the book.

14. Sep 10, 2014

Sounds right