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Given the following part of a circuit, I have to find the susceptance.
I know that the susceptance is \$B=-\frac{1}{reactance}\$.
I know that the resistor does not have a reactance, however, my textbook's solution is \$ B=-\frac{wL}{R^2+(wL)^2} \$. Why does \$R\$ come up in the solution?
Well, the definition of susceptance is the following:
$$\text{susceptance}=\Im\left(\text{admittance}\right)=\Im\left(\frac{1}{\text{impedance}}\right)\tag1$$
In formula form:
$$\text{B}=\Im\left(\underline{\text{Y}}\right)=\Im\left(\frac{1}{\underline{\text{Z}}}\right)\tag2$$
So, we get:
\begin{equation} \begin{split} \text{B}&=\Im\left(\frac{1}{\text{R}+\text{j}\omega\text{L}}\right)\\ \\ &=\Im\left(\frac{1}{\text{R}+\text{j}\omega\text{L}}\cdot\frac{\text{R}-\text{j}\omega\text{L}}{\text{R}-\text{j}\omega\text{L}}\right)\\ \\ &=\Im\left(\frac{\text{R}-\text{j}\omega\text{L}}{\text{R}^2+\left(\omega\text{L}\right)^2}\right)\\ \\ &=\Im\left(\frac{\text{R}}{\text{R}^2+\left(\omega\text{L}\right)^2}-\frac{\omega\text{L}}{\text{R}^2+\left(\omega\text{L}\right)^2}\cdot\text{j}\right)\\ \\ &=\Im\left(\underbrace{\frac{\text{R}}{\text{R}^2+\left(\omega\text{L}\right)^2}}_{=\space\Re}+\left(\underbrace{-\frac{\omega\text{L}}{\text{R}^2+\left(\omega\text{L}\right)^2}}_{=\space\Im}\right)\cdot\text{j}\right)\\ \\ &=-\frac{\omega\text{L}}{\text{R}^2+\left(\omega\text{L}\right)^2} \end{split}\tag3 \end{equation}
Because the total impedance of your circuit is a series circuit of a resistor, with impedance \$\text{R}\$, and an inductor with impedance \$\text{j}\omega\text{L}\$. So:
$$\underline{\text{Z}}=\text{R}+\text{j}\omega\text{L}\tag4$$
