Inverse Tangent of i

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Theorem

The inverse tangent of $i$ is not defined.


Proof

Aiming for a contradiction, suppose $\tan z_0 = i$.

\(\displaystyle \dfrac {\sin z_0} {\cos z_0}\) \(=\) \(\displaystyle i\) Definition of Tangent Function
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sin z_0\) \(=\) \(\displaystyle i \cos z_0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sin^2 z_0\) \(=\) \(\displaystyle -\cos^2 z_0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sin^2 z_0 + \cos^2 z_0\) \(=\) \(\displaystyle 0\)

This contradicts the theorem Sum of Squares of Sine and Cosine:

$\sin^2 z_0 + \cos^2 z_0 = 1$

Hence the result by Proof by Contradiction.

$\blacksquare$


Sources