Can inverse trigonometric functions be complex? Can inverse trigonometric functions be solved by complex? Yes, inverse trigonometric functions can be solved using complex numbers. In complex analysis, the inverse trigonometric functions are defined in a way that extends their usual definition on the real line. Like `\sin^-1 z`, `\cos^-1 z`. `\tan^-1 z` can be solved using complex numbers.
It is important to note that the solutions to inverse trigonometric functions using complex numbers may not always be unique. This is because the complex logarithm is a multi-valued function.
Inverse Trigonometric Functions in
Complex Analysis
The inverse trigonometric functions are multi-valued functions in the complex
plane, meaning that they can take on multiple values for a given input. This is
because trigonometric functions have periodicity in the complex plane, and the
inverse trigonometric functions have to take this into account.
Complex inverse trigonometric functions can be described in terms
of logarithmic functions. Here is some proof of trigonometric functions.
Before starting the proves, we recall the formulas of sinx and cosx.
`\cos x =\frac{e^(ix) + e^(-ix)}{2}`
`\sin x =\frac{e^(ix) - e^(-ix)}{2i}`
Prove of `\sin^-1 z` in terms of logarithm function.
Let us define
w = `\sin^-1 z`
⇒ 𝑧 = sin 𝑤
z = `\frac{e^(iw) - e^(-iw)}{2i}`
2𝑖𝑧 = `e^(iw) - e^(-iw)`
`(e^(iw))^2 − 2ize^(iw) − 1 = 0`
By using the quadratic formula:
`e^(iw) = \frac{-(-2iz) ± \sqrt{(-2iz)^2 -4(1)(-1)}}{2(1)}`
= `\frac{2iz ± \sqrt{4 - 4z^2}}{2}`
= `\frac{2iz ± 2 \sqrt{1 - z^2}}{2}`
= `iz ± \sqrt{1 - z^2}`
Taking '𝑙𝑛' on both sides
`lne^(iw) = ln(iz ± \sqrt{1 - z^2})`
`iw lne = ln(iz ± \sqrt{1 - z^2})`
`iw = ln(iz ± \sqrt{1 - z^2})`
`w = \frac{1}{i} ln(iz ± \sqrt{1 - z^2})`
`\sin^-1 z = -i ln(iz ± \sqrt{1 - z^2})`
Hence proved.
Prove of `\cos^-1 z` in terms of logarithm function.
Let us define
w = `\cos^-1 z`
⇒ 𝑧 = coz 𝑤
z = `\frac{e^(iw) + e^(-iw)}{2}`
2𝑧 = `e^(iw) + e^(-iw)`
`(e^(iw))^2 − 2ze^(iw) + 1 = 0`
By using the quadratic formula:
`e^(iw)` = `\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}`
= `z ± \sqrt{-1(1 - z^2)}`
= `z ± i \sqrt{1 - z^2}`
Taking '𝑙𝑛' on both sides
`lne^(iw) = ln(z ± i \sqrt{1 - z^2})`
`iw lne = ln(z ± i \sqrt{1 - z^2})`
`iw = ln(z ± i \sqrt{1 - z^2})`
`w = \frac{1}{i} ln(z ± i \sqrt{1 - z^2})`
`\cos^-1 z = -i ln(z ± i \sqrt{1 - z^2})`
Hence proved.
Prove of `\tan^-1 z` in terms of logarithm function.
Let us define
w = `\tan^-1 z`
⇒ 𝑧 = tan 𝑤
z = `\frac{1}{i} \frac{e^(iw) - e^(-iw)}{e^(iw) + e^(-iw)}`
`iz = \frac{e^(iw) - e^(-iw)}{e^(iw) + e^(-iw)}`
`iz = \frac{e^(iw)(1 - e^(-2iw))}{e^(iw)(1 + e^(-2iw))}`
`iz = \frac{(1 - e^(-2iw))}{(1 + e^(-2iw))}`
`(1 + e^(-2iw))iz = (1 - e^(-2iw))`
`iz + iz e^(-2iw) = 1 - e^(-2iw)`
`iz + iz e^(-2iw) + e^(-2iw) = 1`
`(1 + iz)e^(-2iw) = 1 - iz`
`e^(-2iw) = \frac{1 - iz}{1 + iz}`
`e^(-2iw) = \frac{i(\frac{1}{i} - z)}{i(\frac{1}{i} + z)}`
`e^(-2iw) = \frac{(-i - z)}{(-i + z)}`
`e^(-2iw) = \frac{-(i + z)}{-(i - z)}`
`e^(-2iw) = \frac{(i + z)}{(i - z)}`
Taking '𝑙𝑛' on both sides
`-2iw = ln(\frac{i + z}{i - z})`
`2iw = ln(\frac{i + z}{i - z})`
`w = \frac{-1}{2i} ln(\frac{i + z}{i - z})`
`w = \frac{i}{2} ln(\frac{i + z}{i - z})`
Now,
`\tan^-1 z = \frac{i}{2} ln(\frac{i + z}{i - z})`
Hence proved.
Derivatives of Inverse Trigonometric Functions
In complex analysis, the derivative of inverse trigonometric functions are defined in a similar way as in real analysis, but using complex numbers instead of real numbers. The formulas for derivatives of inverse trigonometric functions in complex analysis are:
`\frac{d}{dz} \sin^-1 z = \frac{1}{\sqrt{1 - z^2}}`
`\frac{d}{dz} \cos^-1 z = \frac{-1}{\sqrt{1 - z^2}}`
`\frac{d}{dz} \tan^-1 z = \frac{1}{1 - z^2}`
These formulas hold for any complex number z, with the exception of points on the branch cuts of the functions, where they are not defined. The branch cuts of the inverse sine and inverse cosine functions are the interval [- infinity, -1] and [1, infinity] on the real axis and the branch cuts of the inverse tangent function is the interval [`-i, i`] on the imaginary axis.
It is also worth noting that the derivative of inverse trigonometric functions can be derived using complex logarithm differentiation which involves taking the logarithm of both sides of the function and then differentiating with respect to z. However, this approach requires careful consideration of branch cuts or may not always be the most convenient method for computing the derivatives.
Examples:
Example-1: Find all values of `\sin^-1 (-i)`
Solution: Let 𝑧 = −𝑖 = 0 − 1.𝑖
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