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Does anyone know where to start?

I have tried it several times, starting with simplifying the left part by multiplying the conjugate and no luck.

$$ (\frac{1+i p}{1-ip})e^{2i\cot^{-1}p}=1$$

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2 Answers 2

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Actually a minus sign is missing on the right side.

Let $q=\cot^{-1} p$. Then $\frac {1+ip} {1-ip} =\frac {1+i\cot q} {1-i \cot q} =\frac {\sin q +i\cos q} {\sin q -i\cos q}=-e^{-2iq}$. Multiply both sides by $e^{2iq}$. We get $\frac {1+ip} {1-ip} e^{2i\cot ^{-1} p} =-1$.

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Hint:

WLOG

$1=r\cos A,p=r\sin A$

$$\dfrac{1+ip}{1-ip}=\dfrac{\cos A+i\sin A}{\cos A-i\sin A}=\dfrac{e^{iA}}{e^{-iA}}=e^{2iA}$$

$\dfrac p1=\tan A$

$\cot^{-1}p=\dfrac\pi2-\tan^{-1}(\tan A)$

Use atan2

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