Hyperbolic Functions

Work in progress. This note is still being written and incomplete.

For any complex number zz,

sinh⁡z=ez−e−z2,cosh⁡z=ez+e−z2\sinh z = \frac{e^z - e^{-z}}{2}, \qquad \cosh z = \frac{e^z + e^{-z}}{2}

sinh⁡z\sinh z and cosh⁡z\cosh z are entire functions.

Other Hyperbolic Functions

tanh⁡z=sinh⁡zcosh⁡z,coth⁡z=cosh⁡zsinh⁡z\tanh z = \frac{\sinh z}{\cosh z}, \qquad \coth z = \frac{\cosh z}{\sinh z} sech⁡z=1cosh⁡z,csch⁡z=1sinh⁡z\operatorname{sech} z = \frac{1}{\cosh z}, \qquad \operatorname{csch} z = \frac{1}{\sinh z}

Analyticity

sinh⁡z\sinh z and cosh⁡z\cosh z are analytic on C\mathbb{C}, since eze^z and e−ze^{-z} are.

The other 4 functions are quotients, so each is analytic wherever its denominator is non-zero.

FunctionAnalytic on
tanh⁡z\tanh z, sech⁡z\operatorname{sech} zC∖{i(π2+nπ)}\mathbb{C} \setminus \{i(\frac{\pi}{2} + n\pi)\}
coth⁡z\coth z, csch⁡z\operatorname{csch} zC∖{nπi}\mathbb{C} \setminus \{n\pi i\}

Here n∈Zn \in \mathbb{Z}.

sinh⁡z=0\sinh z = 0 iff z=nπiz = n\pi i, and cosh⁡z=0\cosh z = 0 iff z=i(π2+nπ)z = i(\frac{\pi}{2} + n\pi). All zeros are purely imaginary.

Derivatives

  • ddz cosh⁡z=sinh⁡z\dfrac{\text{d}}{\text{d}z}\,\cosh z = \sinh z
  • ddz sinh⁡z=cosh⁡z\dfrac{\text{d}}{\text{d}z}\,\sinh z = \cosh z
  • ddz tanh⁡z=sech⁡2z\dfrac{\text{d}}{\text{d}z}\,\tanh z = \operatorname{sech}^2 z
  • ddz coth⁡z=−csch⁡2z\dfrac{\text{d}}{\text{d}z}\,\coth z = -\operatorname{csch}^2 z
  • ddz sech⁡z=−sech⁡ztanh⁡z\dfrac{\text{d}}{\text{d}z}\,\operatorname{sech} z = -\operatorname{sech} z \tanh z
  • ddz csch⁡z=−csch⁡zcoth⁡z\dfrac{\text{d}}{\text{d}z}\,\operatorname{csch} z = -\operatorname{csch} z \coth z

Identities

Valid whenever the expressions involved are defined.

  • cosh⁡2z−sinh⁡2z=1\cosh^2 z - \sinh^2 z = 1
  • sinh⁡(z1+z2)=sinh⁡z1cosh⁡z2+cosh⁡z1sinh⁡z2\sinh(z_1 + z_2) = \sinh z_1 \cosh z_2 + \cosh z_1 \sinh z_2
  • cosh⁡(z1+z2)=cosh⁡z1cosh⁡z2+sinh⁡z1sinh⁡z2\cosh(z_1 + z_2) = \cosh z_1 \cosh z_2 + \sinh z_1 \sinh z_2
  • cosh⁡(−z)=cosh⁡z\cosh(-z) = \cosh z
  • sinh⁡(−z)=−sinh⁡z\sinh(-z) = -\sinh z
  • cosh⁡(z+2πi)=cosh⁡z\cosh(z + 2\pi i) = \cosh z
  • sinh⁡(z+2πi)=sinh⁡z\sinh(z + 2\pi i) = \sinh z

For z=x+iyz = x + iy:

  • cosh⁡z=cosh⁡xcos⁡y+isinh⁡xsin⁡y\cosh z = \cosh x \cos y + i\sinh x \sin y
  • sinh⁡z=sinh⁡xcos⁡y+icosh⁡xsin⁡y\sinh z = \sinh x \cos y + i\cosh x \sin y

Infinite Series

Both series are Maclaurin series. They converge for all z∈Cz \in \mathbb{C}, since sinh⁡z\sinh z and cosh⁡z\cosh z are entire.

sinh⁡z=∑n=0∞z2n+1(2n+1)!=z+z33!+z55!+⋯\sinh z = \sum_{n=0}^{\infty} \frac{z^{2n+1}}{(2n+1)!} = z + \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots cosh⁡z=∑n=0∞z2n(2n)!=1+z22!+z44!+⋯\cosh z = \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!} = 1 + \frac{z^2}{2!} + \frac{z^4}{4!} + \cdots

Relation to Trigonometric Functions

  • cosh⁡(iz)=cos⁡z\cosh(iz) = \cos z
  • sinh⁡(iz)=isin⁡z\sinh(iz) = i\sin z
  • cos⁡(iz)=cosh⁡z\cos(iz) = \cosh z
  • sin⁡(iz)=isinh⁡z\sin(iz) = i\sinh z
Written by October 17, 2026 1 min read
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