The complex inverse hyperbolic functions are the inverse functions of the hyperbolic functions in the complex plane. They are multi-valued functions that can have multiple branches, and their values depend on the choice of branch. The complex inverse hyperbolic functions include the complex inverse hyperbolic sine (arcsinh or asinh), complex inverse hyperbolic cosine (arccosh or acosh), complex inverse hyperbolic tangent (arctanh or atanh), complex inverse hyperbolic cotangent (arcoth or acoth), complex inverse hyperbolic secant (arcsech or asech), and complex inverse hyperbolic cosecant (arccsch or acsch).

The complex inverse hyperbolic functions have applications in complex analysis, physics and engineering, such as in the study of complex functions, special relativity, quantum mechanics, analytic continuation and conformal mapping.

The complex inverse hyperbolic functions are defined in the same way as the real inverse hyperbolic functions, but the logarithm used is the complex logarithm function.

Complex Inverse Hyperbolic Functions

Complex inverse hyperbolic functions can also be described in terms of logarithmic functions. Here is some proof of complex inverse hyperbolic functions.

Before starting the proves, we recall the formulas of sinhx and coshx.

`\sinh z =\frac{e^z - e^(-z)}{2}`          

`\cosh z =\frac{e^z + e^(-z)}{2}`

Prove of  `\sinh^-1 z` in terms of logarithm function.

Let us define

w = `\sinh^-1 z`

⇒ 𝑧 = sinh 𝑤

z = `\frac{e^w - e^(-w)}{2}`

2𝑧 = `e^(w) - e^(-w)`

`(e^w)^2 − 2ze^(w) − 1 = 0`

By using the quadratic formula:

`e^w = \frac{-(-2z) ± \sqrt{(-2z)^2 -4(1)(-1)}}{2(1)}`

= `\frac{2z ± \sqrt{4z^2 + 4}}{2}`

= `\frac{2z ± 2 \sqrt{z^2 + 1}}{2}`

= `z ± \sqrt{z^2 + 1}`

Taking '𝑙𝑛' on both sides

`lne^(w) = ln(z ± \sqrt{z^2 + 1})`

`w = ln(z ± \sqrt{z^2 + 1})`

 `\sinh^-1 z = ln(z ± \sqrt{z^2 + 1})`

Hence proved.

Prove of  `\cosh^-1 z` in terms of logarithm function.

Let us define

w = `\cosh^-1 z`

⇒ 𝑧 = cosh 𝑤

z = `\frac{e^w + e^(-w)}{2}`

2𝑧 = `e^w + e^(-w)`

`(e^w)^2 − 2ze^w + 1 = 0`

By using the quadratic formula:

`e^w` = `\frac{-(-2z) ± \sqrt{(-2z)^2 -4(1)(1)}}{2(1)}`

= `\frac{2z ± \sqrt{4z^2 -4}}{2}`

= `\frac{2z ± 2 \sqrt{z^2 -1}}{2}`

= `z ± \sqrt{z^2 -1}`

Taking '𝑙𝑛' on both sides

`lne^(w) = ln(z ± \sqrt{z^2 - 1})`

`w = ln(z ± \sqrt{z^2 - 1})`

 `\cosh^-1 z = ln(z ± \sqrt{z^2 - 1})`

Hence proved.

Prove of  `\tanh^-1 z` in terms of logarithm function.

Let us define

w = `\tanh^-1 z`

⇒ 𝑧 = tanh 𝑤

`z = \frac{e^(w) - e^(-w)}{e^(w) + e^(-w)}`

z = `\frac{e^(-w)(e^(2w) - 1)}{e^(-w)( e^(2w) + 1)}`

z = `\frac{(e^(2w) - 1)}{( e^(2w) + 1)}`

`( e^(2w) + 1)z = e^(2w) - 1`

`e^(2w) z + z = e^(2w) - 1`

`e^(2w) z - e^(2w)  =  - 1 - z`

Multiply minus on b/s

`e^(2w) - e^(2w) z = 1 + z`

`e^(2w)(1 - z) = 1 + z`

`e^(2w) = \frac{1 + z}{1 - z}`

Taking '𝑙𝑛' on both sides

`2w = ln(\frac{1 + z}{1 - z})`

`w = \frac{1}{2}ln(\frac{1 + z}{1 - z})`

`\tanh^-1 z = \frac{1}{2}ln(\frac{1 + z}{1 - z})`

Hence proved.

Complex Trigonometric and Hyperbolic Functions are related. 

If we replace z with iz and then, we obtain:

`\cosh iz = \cos z                       \sinh iz = i sinz`

Example:1

Find all values of  `\tanh^-1 (0)`

Solution: Let 𝑧 = 0 = 0 + 1.`i`

`\tanh^-1 z = \frac{1}{2}ln(\frac{1 + z}{1 - z})`

`\tanh^-1 (0) = \frac{1}{2}ln(\frac{1 + 0}{1 - 0})`

`\tanh^-1 (0) = \frac{1}{2}ln(1)`

`\tanh^-1 (0) = \frac{1}{2}`[ln(1) + 2𝑛𝜋𝑖]        𝑛 = 0, ±1, ±2, ±3, …

`\tanh^-1 (0) = \frac{1}{2}`[0 + 2𝑛𝜋𝑖]             𝑛 = 0, ±1, ±2, ±3, …

`\tanh^-1 (0)` = 𝑛𝜋𝑖                              𝑛 = 0, ±1, ±2, ±3, …

Example:2

Find all values of `\cosh^-1 (-1)`

Solution: Let 𝑧 = −1 = 0 − 1.`i`

`\cosh^-1 z = ln(z ± \sqrt{z^2 - 1})`

`\cosh^-1 (-1) = ln(-1 ± \sqrt{(-1)^2 - 1})`

`\cosh^-1 (-1) = ln(-1)`

`\cosh^-1 (-1)` = ln|−1| + 𝜋𝑖 + 2𝑛𝜋𝑖                          𝑛 = 0, ±1, ±2, ±3, …

`\cosh^-1 (-1)` = (2𝑛 + 1)𝜋𝑖                                      𝑛 = 0, ±1, ±2, ±3, …

Example:3

Find all values of  `\sinh^-1 (-i)

Solution: Let 𝑧 = −𝑖 = 0 − 1.𝑖

`\sinh^-1 z = ln(z ± \sqrt{z^2 + 1})`

 `\sinh^-1 (-i) = ln(-i ± \sqrt{(-i)^2 + 1})`

`\sinh^-1 (-i) = ln(-i ± \sqrt{-1 + 1})`

`\sinh^-1 (-i) = ln(-i)`

`\sinh^-1 (-i) = ln|-1| - \frac{\pi}{2}` + 2𝑛𝜋𝑖       𝑛 = 0, ±1, ±2, ±3, …

`\sinh^-1 (-i) = 0 - \frac{\pi}{2}` + 2𝑛𝜋𝑖             𝑛 = 0, ±1, ±2, ±3, …

`\sinh^-1 (-i)` = `(2n - \frac{1}{2})`𝜋𝑖                𝑛 = 0, ±1, ±2, ±3, …

Practice Questions

Q-1: Find all values of the following

a) `\sinh^-1 (2i)`

b) `\cosh^-1 (2i)`

c) `\tanh^-1 (1 + i)`

Q-2: Solve the equation 𝑠𝑖𝑛ℎ 𝑧 = 2 for ‘z’.