How do I see $\bigg| (e^{2\pi i\xi}-\lambda)(e^{2\pi i\xi}-\frac{1}{\lambda}) \bigg| = \bigg|2\cos(2\pi\xi)-\frac{1}{\lambda}+\lambda\bigg|$?

Question

$$\bigg| \left(e^{2\pi i\xi}-\lambda\right)\left(e^{2\pi i\xi}-\frac{1}{\lambda}\right) \bigg| = \bigg|2\cos(2\pi\xi)-\frac{1}{\lambda}+\lambda\bigg|$$

Here $i^2 = -1$, and $\lambda<0$ is a constant.

Answer 1

$$\bigg| (e^{2\pi i\xi}-\lambda)(e^{2\pi i\xi}-\frac{1}{\lambda}) \bigg|^2= (e^{2\pi i\xi}-\lambda)(e^{-2\pi i\xi}-\lambda)(e^{2\pi i\xi}-\frac{1}{\lambda})(e^{-2\pi i\xi}-\frac{1}{\lambda})\\ =(1-\lambda(e^{2\pi i\xi}+e^{-2\pi i\xi})+\lambda^2)(1-\frac{1}{\lambda}(e^{2\pi i\xi}+e^{-2\pi i\xi})+\frac{1}{\lambda^2})\\ =(1-\lambda(2\cos{2\pi\xi})+\lambda^2)(1-\frac{1}{\lambda}(2\cos{2\pi\xi})+\frac{1}{\lambda^2})\\ \implies\bigg| (e^{2\pi i\xi}-\lambda)(e^{2\pi i\xi}-\frac{1}{\lambda}) \bigg|=\bigg|\frac{1-2\lambda\cos{2\pi\xi}+\lambda^2}{\lambda}\bigg|=\bigg|2\cos(2\pi\xi)-\frac{1}{\lambda}-\lambda\bigg|$$