Impedance network and complex algebra

[cjs94]

Homework Statement

This is one part of a wider question, I'm only posting the part I'm having trouble with.
$$
\begin{align}
\text{Given an impedance network } B &= \frac{Z_1 \parallel Z_3}{Z_2 + Z_1 \parallel Z_3} \\
\text{show that: } \frac{1}{B} &= 1 + \frac{R_2}{R_1} + j\frac{\omega CR_2}{1 + \omega CR_3}
\end{align}
$$

Homework Equations

$$\begin{align}
Z_1 &= R_1 \\
Z_2 &= R_2 \\
Z_3 &= R_3 + C_1 \text{(in series)}
\end{align}$$

The Attempt at a Solution

$$\begin{align}
Z_3 &= R_3 - j\frac{1}{\omega C_1} \\
Z_1 \parallel Z_3 &= \frac{Z_1 Z_3}{Z_1 + Z_3} \\
&= \frac{R_1 \left( R_3 - j\frac{1}{\omega C_1} \right)}{R_1 + R_3 - j\frac{1}{\omega C_1}} \\
\frac{1}{B} &= \frac{Z_1 \parallel Z_3}{Z_1 \parallel Z_3} + \frac{Z_2}{Z_1 \parallel Z_3} \\
&= 1 + \frac{R_2}{\frac{R_1 \left( R_3 - j\frac{1}{\omega C_1} \right)}{R_1 + R_3 - j\frac{1}{\omega C_1}}} \\
&= 1 + \frac{R_2 \left( R_1 + R_3 - j\frac{1}{\omega C_1} \right)}{R_1 \left( R_3 - j\frac{1}{\omega C_1} \right)}\\
&= 1 + \frac{R_2 \left( R_3 - j\frac{1}{\omega C_1} \right)}{R_1 \left( R_3 - j\frac{1}{\omega C_1} \right)} + \frac{R_2 R_1}{R_1 \left( R_3 - j\frac{1}{\omega C_1} \right)}\\
&= 1 + \frac{R_2}{R_1} + \frac{R_2}{R_3 - j\frac{1}{\omega C_1}} \\
&= 1 + \frac{R_2}{R_1} + \frac{j\omega C_1 R_2}{1+ j\omega C_1 R_3}
\end{align}$$
I just can't get the right hand element to match, I've got an extra $j$. What am I doing wrong?

 

[NascentOxygen]

It's a fair bet that you are right, and the typesetter has dropped an italic j.

I followed your working, nothing leapt out as being wrong.

One way to check your own work is to assign the pronumerals some easy numeric values, and evaluate the expressions (manually, or using a calculator).
e.g., let R1=1, R2=4, R3=3, C=5, w=1

 

[NascentOxygen]

You can use wolframalpha to simplify your complex arithmetic expressions, e.g., http://m.wolframalpha.com/input/?i=1+(1+i4)/(2-i6)&x=0&y=0

I swapped j for the mathematicians i to get it to do what I wanted.

 

[cjs94]

Thanks for the replies. I've just had a response from the tutor, the question is wrong and my answer is correct. :)