Inverse Tangent of i
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Theorem
The inverse tangent of $i$ is not defined.
Proof
Aiming for a contradiction, suppose $\tan z_0 = i$.
\(\ds \dfrac {\sin z_0} {\cos z_0}\) | \(=\) | \(\ds i\) | Definition of Tangent Function | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sin z_0\) | \(=\) | \(\ds i \cos z_0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sin^2 z_0\) | \(=\) | \(\ds -\cos^2 z_0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sin^2 z_0 + \cos^2 z_0\) | \(=\) | \(\ds 0\) |
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
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 4.6$. The Logarithm: Examples: $\text {(iv)}$