Euler's Tangent Identity/Formulation 1

Theorem

Let $z$ be a complex number.

Let $\tan z$ denote the tangent function and $i$ denote the imaginary unit: $i^2 = -1$.

Then:

$\tan z = i \dfrac {1 - e^{2 i z} } {1 + e^{2 i z} }$

Proof

\(\ds \tan z\)

\(=\)

\(\ds \frac {\sin z} {\cos z}\)

Definition of Complex Tangent Function

\(\ds \)

\(=\)

\(\ds \frac {\frac 1 2 i \paren {e^{-i z} - e^{i z} } } {\frac 1 2 \paren {e^{-i z} + e^{i z} } }\)

Euler's Sine Identity and Euler's Cosine Identity

\(\ds \)

\(=\)

\(\ds i \frac {e^{-i z} - e^{i z} } {e^{-i z} + e^{i z} }\)

\(\ds \)

\(=\)

\(\ds i \frac {1 - e^{2 i z} } {1 + e^{2 i z} }\)

multiplying numerator and denominator by $e^{i z}$

$\blacksquare$