I am working through a calculation and it says that when $\omega$ is close to $\omega_L$, the second term below describes a circle with a diameter of $d = \frac{2 \kappa}{1 + \kappa}$. Can someone help me see why this is true?
$$\Gamma=\Gamma_0 \left(1-\frac{2\kappa}{1+\kappa}\frac{1}{1+i Q_L 2 \frac{\omega - \omega_L}{\omega_0}} \right)$$
Let $$\theta=Q_L2\frac{\omega-\omega_L}{\omega_0}<<1$$ $$\frac{1}{1+i\theta}=\frac{1-i\theta}{1+\theta^2}\approx 1-i\theta\approx \cos\theta-i\sin\theta=e^{-i\theta}$$ $$\therefore \frac{2\kappa}{1+\kappa}\frac{1}{1+i\theta}\approx \frac{2\kappa}{1+\kappa}e^{-i\theta}$$