Remark

Tag 089Z, permanent. Cite it as [Bourbaki, Tag 089Z].

For complex $z$ one sometimes writes

\cosh z = \frac{1}{2} (e^z + e^{-z}) = \cos iz \\ \sinh z = \frac{1}{2} (e^z - e^{-z}) = -i \sin iz.

These functions thus extend to $\mathbf{C}$ the hyperbolic functions defined on $\mathbf{R}$ .

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Nothing.