Solving the Dirichilet problem in a strip relating to conformal mappings

Question

Observe that if $$e^{i\phi}=\frac{i-e^{\pi t}}{i+e^{\pi t}}$$ then take the imaginary part and differentiate both sides to establsh the two following identities: $$\sin{\phi}=\frac{1}{\cosh{\pi t}},\ \frac{d\phi}{dt}=\frac{\pi}{\cosh{\pi t}}.$$ I had successfully constructed the second identity by differentiating, but I don't know how to derive the second one.

Answer 1

Rearrange the quantity on the right by multiplying the top and bottom by the complex conjugate of the bottom:

$$\frac{i-e^{\pi t}}{i + e^{\pi t}} \cdot \frac{-i+e^{\pi t}}{-i + e^{\pi t}} = \frac{(1-e^{2\pi t}) + i(2e^{\pi t})}{1 + e^{2\pi t}}$$

$$\implies \sin\phi = \text{Im}\left(\frac{i-e^{\pi t}}{i + e^{\pi t}}\right) = \frac{2e^{\pi t}}{1 + e^{2\pi t}} = \frac{2}{e^{-\pi t} + e^{\pi t}} \equiv \text{sech}(\pi t)$$

As a bonus, using

$$\cos\phi = \text{Re}\left(\frac{i-e^{\pi t}}{i + e^{\pi t}}\right) = \frac{1-e^{2\pi t}}{1 + e^{2\pi t}} = \frac{e^{-\pi t}-e^{\pi t}}{e^{\pi t} + e^{\pi t}} \equiv -\tanh(\pi t)$$

we get that

$$\frac{d}{dt}(\cos \phi) = -\pi \hspace{2 pt} \text{sech}^2(\pi t) \implies -\sin\phi \frac{d\phi}{dt} = -\text{sech}(\pi t) \frac{d\phi}{dt} = -\pi \hspace{2 pt} \text{sech}^2(\pi t)$$

$$\therefore \frac{d\phi}{dt} = \pi \hspace{2 pt}\text{sech}(\pi t)$$