Key to "Proving i2lsnx+ix−i=arctan⁡x"

Adela Brown

Answered question

2022-01-03

Proving

\frac{i}{2} l s n \frac{x + i}{x - i} = \arctan x

Answer & Explanation

usaho4w

Beginner2022-01-04Added 39 answers

HINT:
Let

\arctan x = y \Rightarrow x = \tan y

\frac{x + i}{x - i} = \frac{\sin y + i \cos y}{\sin y - i \cos y} = \frac{i (\cos y - i \sin y)}{- i (\cos y + i \sin y)} = - e^{- 2 i y}

John Koga

Beginner2022-01-05Added 33 answers

I dont

Vasquez

Skilled2022-01-11Added 457 answers

Identity is :
$\arctan z = \frac{i}{2} \ln \frac{i + z}{i - z}$
Proof :
$\tan z = \frac{e^{i z} - e^{- i z}}{i (e^{i z} + e^{- i z})} e^{i \frac{i}{2} \ln \frac{i + z}{i - z}} = (\frac{i + z}{i - z})^{- \frac{1}{2}} e^{- i \frac{i}{2} \ln \frac{i + z}{i - z}} = (\frac{i + z}{i - z})^{\frac{1}{2}} \tan (\frac{i}{2} \ln \frac{i + z}{i - z}) = \frac{1 - \frac{i + z}{i - z}}{i (1 + \frac{i + z}{i - z})} = z$

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