Hyperbolic Functions Notes FE Sem 1 Applied Maths-I | Mumbai Univ

This module extends sine, cosine and the exponential function to complex arguments and introduces the hyperbolic functions sinh, cosh and tanh that appear naturally when the angle is imaginary. You will learn the identities connecting circular and hyperbolic functions, the logarithmic forms of the inverse hyperbolic functions, how to split any such function of x + iy into real and imaginary parts, and how to find the logarithm of a complex number. These tools are used in transmission line theory, AC circuits and electromagnetic waves.

Summary

Circular functions of a complex number (2.1): from Euler's formula, for any complex z, \( \cos z = \frac{e^{iz} + e^{-iz}}{2} \) and \( \sin z = \frac{e^{iz} - e^{-iz}}{2i} \); tan z, cot z, sec z and cosec z are defined from these as usual. All ordinary trigonometric identities continue to hold for complex z.

Hyperbolic functions (2.1): for real or complex x, \( \sinh x = \frac{e^x - e^{-x}}{2} \), \( \cosh x = \frac{e^x + e^{-x}}{2} \), \( \tanh x = \frac{\sinh x}{\cosh x} \), with coth, sech and cosech as reciprocals. sinh and tanh are odd, cosh is even, and \( \cosh x + \sinh x = e^x \). Basic identities are \( \cosh^2x - \sinh^2x = 1 \), \( \text{sech}^2x + \tanh^2x = 1 \), \( \coth^2x - \text{cosech}^2x = 1 \), \( \sinh2x = 2\sinh x\cosh x \) and \( \cosh2x = \cosh^2x + \sinh^2x = 2\cosh^2x - 1 = 1 + 2\sinh^2x \).

Relations between circular and hyperbolic functions: \( \sin(ix) = i\sinh x \), \( \cos(ix) = \cosh x \), \( \tan(ix) = i\tanh x \), and conversely \( \sinh(ix) = i\sin x \), \( \cosh(ix) = \cos x \), \( \tanh(ix) = i\tan x \). These let any trigonometric identity be converted into a hyperbolic one (Osborn's rule: change sign wherever a product of two sinh terms appears).

Inverse circular and inverse hyperbolic functions (2.1): if \( \sinh y = x \) then \( y = \sinh^{-1}x \), and solving the quadratic in \( e^y \) gives \( \sinh^{-1}x = \log\left(x + \sqrt{x^2 + 1}\right) \), \( \cosh^{-1}x = \log\left(x + \sqrt{x^2 - 1}\right) \) (x ≥ 1) and \( \tanh^{-1}x = \frac12\log\frac{1 + x}{1 - x} \) (|x| < 1). Inverse circular functions of complex arguments are linked to them, e.g. \( \sin^{-1}(ix) = i\sinh^{-1}x \) and \( \tan^{-1}(ix) = i\tanh^{-1}x \).

Separation of real and imaginary parts (2.1): using the addition formulas with \( \cos iy = \cosh y \), \( \sin iy = i\sinh y \): \( \sin(x + iy) = \sin x\cosh y + i\cos x\sinh y \), \( \cos(x + iy) = \cos x\cosh y - i\sin x\sinh y \), \( \sinh(x + iy) = \sinh x\cos y + i\cosh x\sin y \), \( \cosh(x + iy) = \cosh x\cos y + i\sinh x\sin y \). For tan and tanh multiply numerator and denominator by the conjugate of the denominator to get \( \tan(x + iy) = \frac{\sin2x + i\sinh2y}{\cos2x + \cosh2y} \) and \( \tanh(x + iy) = \frac{\sinh2x + i\sin2y}{\cosh2x + \cos2y} \). A common exam pattern: given \( \tan(\alpha + i\beta) = x + iy \), use \( \tan(\alpha - i\beta) = x - iy \) and add or subtract the angles.

Logarithm of a complex number (2.2): writing \( z = x + iy = re^{i(\theta + 2n\pi)} \), the general logarithm is \( \text{Log}\,z = \log r + i(\theta + 2n\pi) \), n an integer. The principal value (n = 0) is \( \log z = \frac12\log(x^2 + y^2) + i\theta \) where θ is the principal argument (\( \theta = \tan^{-1}(y/x) \) when x > 0). Thus Log z is many-valued; for example \( \log(-1) = i\pi \) and \( \log i = i\pi/2 \) (principal values).

Complex powers: for complex a and b, \( a^b = e^{b\,\text{Log}\,a} \), which is in general many-valued. For example \( i^i = e^{-(4n + 1)\pi/2} \), all of which are real, with principal value \( e^{-\pi/2} \). To separate real and imaginary parts of \( a^b \), find \( b\log a = A + iB \) and write \( a^b = e^A(\cos B + i\sin B) \).

Self-learning topic: applications of complex numbers in electrical circuits, where sinusoidal voltages and currents are represented as phasors and impedances \( Z = R + iX \) are combined using complex algebra; hyperbolic functions appear in the voltage and current equations of long transmission lines.

Mind map

Method: separating real and imaginary parts of tan(x + iy)

Important definitions

Formulas

Key points to remember

• cosh x ≥ 1 for all real x, while sinh x and tanh x take negative values; |tanh x| < 1.
• cos(ix) = cosh x is real; sin(ix) = i sinh x is purely imaginary.
• Osborn's rule: a trig identity becomes a hyperbolic identity if every sin is replaced by sinh and cos by cosh, and the sign of each term containing a product of two sines is changed.
• For problems of the form f(α + iβ) = x + iy, write the conjugate equation f(α − iβ) = x − iy and add, subtract or multiply.
• To eliminate α, use sin² + cos² = 1; to eliminate β, use cosh² − sinh² = 1.
• The logarithm of a complex number has infinitely many values differing by 2πi; the principal value uses the principal argument.
• i^i has infinitely many values, all real.
• log x in this module means the natural logarithm (base e).

Solved examples

Example 1. If \( \tan(\alpha + i\beta) = x + iy \), prove that \( x^2 + y^2 + 2x\cot2\alpha = 1 \).

Given \( \tan(\alpha + i\beta) = x + iy \); taking the conjugate, \( \tan(\alpha - i\beta) = x - iy \).

\( \tan2\alpha = \tan[(\alpha + i\beta) + (\alpha - i\beta)] = \frac{(x + iy) + (x - iy)}{1 - (x + iy)(x - iy)} = \frac{2x}{1 - x^2 - y^2} \).

So \( 1 - x^2 - y^2 = 2x\cot2\alpha \), i.e. \( x^2 + y^2 + 2x\cot2\alpha = 1 \).

Example 2. If \( \sin(\theta + i\phi) = \tan\alpha + i\sec\alpha \), prove that \( \cos2\theta\cosh2\phi = 3 \).

\( \sin(\theta + i\phi) = \sin\theta\cosh\phi + i\cos\theta\sinh\phi \). Equating parts: \( \sin\theta\cosh\phi = \tan\alpha \), \( \cos\theta\sinh\phi = \sec\alpha \).

Square and subtract (sec²α − tan²α = 1): \( \cos^2\theta\sinh^2\phi - \sin^2\theta\cosh^2\phi = 1 \).

Let \( C = \cos2\theta \), \( H = \cosh2\phi \). Then \( \cos^2\theta = \frac{1 + C}{2} \), \( \sin^2\theta = \frac{1 - C}{2} \), \( \sinh^2\phi = \frac{H - 1}{2} \), \( \cosh^2\phi = \frac{H + 1}{2} \).

\( \frac{(1 + C)(H - 1) - (1 - C)(H + 1)}{4} = 1 \Rightarrow \frac{2CH - 2}{4} = 1 \Rightarrow CH = 3 \).

Hence \( \cos2\theta\cosh2\phi = 3 \).

Example 3. If \( \cosh u = \sec\theta \), prove that \( u = \log\tan\left(\frac\pi4 + \frac\theta2\right) \).

\( \sinh u = \sqrt{\cosh^2u - 1} = \sqrt{\sec^2\theta - 1} = \tan\theta \) (taking u and θ positive).

\( e^u = \cosh u + \sinh u = \sec\theta + \tan\theta = \frac{1 + \sin\theta}{\cos\theta} \).

Using half angles: \( \frac{1 + \sin\theta}{\cos\theta} = \frac{(\cos\frac\theta2 + \sin\frac\theta2)^2}{\cos^2\frac\theta2 - \sin^2\frac\theta2} = \frac{\cos\frac\theta2 + \sin\frac\theta2}{\cos\frac\theta2 - \sin\frac\theta2} = \frac{1 + \tan\frac\theta2}{1 - \tan\frac\theta2} = \tan\left(\frac\pi4 + \frac\theta2\right) \).

Taking logarithms, \( u = \log\tan\left(\frac\pi4 + \frac\theta2\right) \).

Example 4. Separate into real and imaginary parts: \( \tanh(x + iy) \).

Multiply numerator and denominator by \( 2\cosh(x - iy) \): \( \tanh(x + iy) = \frac{2\sinh(x + iy)\cosh(x - iy)}{2\cosh(x + iy)\cosh(x - iy)} \).

Using \( 2\sinh A\cosh B = \sinh(A + B) + \sinh(A - B) \): numerator = \( \sinh2x + \sinh(2iy) = \sinh2x + i\sin2y \).

Using \( 2\cosh A\cosh B = \cosh(A + B) + \cosh(A - B) \): denominator = \( \cosh2x + \cosh(2iy) = \cosh2x + \cos2y \).

Real part \( = \frac{\sinh2x}{\cosh2x + \cos2y} \), imaginary part \( = \frac{\sin2y}{\cosh2x + \cos2y} \).

Example 5. Show that all values of \( i^i \) are real and find the principal value.

\( i = \cos\frac\pi2 + i\sin\frac\pi2 \Rightarrow \text{Log}\,i = i\left(2n\pi + \frac\pi2\right) \).

\( i^i = e^{i\,\text{Log}\,i} = e^{i\cdot i(2n\pi + \pi/2)} = e^{-(4n + 1)\pi/2} \).

Each value is a real exponential, so all values are real. Principal value (n = 0): \( e^{-\pi/2} \approx 0.2079 \).

Example 6. Find the principal value of \( (1 + i)^{1 - i} \) in the form a + ib.

\( \log(1 + i) = \frac12\log2 + i\frac\pi4 \).

\( (1 - i)\log(1 + i) = \frac12\log2 + i\frac\pi4 - \frac i2\log2 + \frac\pi4 = \left(\frac12\log2 + \frac\pi4\right) + i\left(\frac\pi4 - \frac12\log2\right) \).

\( (1 + i)^{1 - i} = e^{\frac12\log2 + \frac\pi4}\, e^{i(\frac\pi4 - \frac12\log2)} = \sqrt2\,e^{\pi/4}\left[\cos\left(\frac\pi4 - \frac12\log2\right) + i\sin\left(\frac\pi4 - \frac12\log2\right)\right] \).

Example 7. Prove that \( \tanh^{-1}x = \frac12\log\frac{1 + x}{1 - x} \) and hence show \( \tan^{-1}\left[i\,\frac{x - a}{x + a}\right] = \frac i2\log\frac xa \).

Let \( y = \tanh^{-1}x \), so \( x = \frac{e^y - e^{-y}}{e^y + e^{-y}} = \frac{e^{2y} - 1}{e^{2y} + 1} \).

Solving: \( e^{2y} = \frac{1 + x}{1 - x} \), so \( y = \frac12\log\frac{1 + x}{1 - x} \).

Now \( \tan(i\theta) = i\tanh\theta \Rightarrow \tan^{-1}(iz) = i\tanh^{-1}z \).

With \( z = \frac{x - a}{x + a} \): \( \frac{1 + z}{1 - z} = \frac{2x}{2a} = \frac xa \).

So \( \tan^{-1}\left[i\frac{x - a}{x + a}\right] = i\cdot\frac12\log\frac xa = \frac i2\log\frac xa \).

Example 8. If \( 5\sinh x - \cosh x = 5 \), find \( \tanh x \).

\( \frac52(e^x - e^{-x}) - \frac12(e^x + e^{-x}) = 5 \Rightarrow 2e^x - 3e^{-x} = 5 \).

Multiply by \( e^x \) (put \( t = e^x \)): \( 2t^2 - 5t - 3 = 0 \Rightarrow (2t + 1)(t - 3) = 0 \).

Since \( e^x > 0 \), t = 3, so x = log 3.

\( \tanh x = \frac{e^{2x} - 1}{e^{2x} + 1} = \frac{9 - 1}{9 + 1} = \frac45 \).

Important questions and answers

Q1. Prove that \( \sinh^{-1}x = \log\left(x + \sqrt{x^2 + 1}\right) \). (3 marks)

Let \( y = \sinh^{-1}x \), so \( x = \sinh y = \frac{e^y - e^{-y}}{2} \).

\( e^{2y} - 2xe^y - 1 = 0 \Rightarrow e^y = x \pm \sqrt{x^2 + 1} \).

Since \( e^y > 0 \) and \( x - \sqrt{x^2 + 1} < 0 \), take the positive sign: \( y = \log\left(x + \sqrt{x^2 + 1}\right) \).

Q2. If \( \sin(\alpha + i\beta) = x + iy \), prove that \( \frac{x^2}{\cosh^2\beta} + \frac{y^2}{\sinh^2\beta} = 1 \) and \( \frac{x^2}{\sin^2\alpha} - \frac{y^2}{\cos^2\alpha} = 1 \). (5 marks)

\( \sin(\alpha + i\beta) = \sin\alpha\cosh\beta + i\cos\alpha\sinh\beta \), so \( x = \sin\alpha\cosh\beta \), \( y = \cos\alpha\sinh\beta \).

Eliminating α: \( \frac{x^2}{\cosh^2\beta} + \frac{y^2}{\sinh^2\beta} = \sin^2\alpha + \cos^2\alpha = 1 \).

Eliminating β: \( \frac{x^2}{\sin^2\alpha} - \frac{y^2}{\cos^2\alpha} = \cosh^2\beta - \sinh^2\beta = 1 \).

Q3. Separate into real and imaginary parts: \( \cos(x + iy) \) and \( \sinh(x + iy) \). (3 marks)

\( \cos(x + iy) = \cos x\cos iy - \sin x\sin iy = \cos x\cosh y - i\sin x\sinh y \).

\( \sinh(x + iy) = \sinh x\cosh iy + \cosh x\sinh iy = \sinh x\cos y + i\cosh x\sin y \).

Q4. If \( \tan\frac x2 = \tanh\frac y2 \), prove that \( \cosh y = \sec x \). (4 marks)

Let \( t = \tan\frac x2 = \tanh\frac y2 \).

\( \cos x = \frac{1 - \tan^2\frac x2}{1 + \tan^2\frac x2} = \frac{1 - t^2}{1 + t^2} \).

\( \cosh y = \frac{1 + \tanh^2\frac y2}{1 - \tanh^2\frac y2} = \frac{1 + t^2}{1 - t^2} \).

Hence \( \cosh y = \frac{1}{\cos x} = \sec x \).

Q5. Prove that \( \log\frac{a + ib}{a - ib} = 2i\tan^{-1}\frac ba \) (principal values, a > 0). (3 marks)

\( \log(a + ib) = \frac12\log(a^2 + b^2) + i\tan^{-1}\frac ba \), \( \log(a - ib) = \frac12\log(a^2 + b^2) - i\tan^{-1}\frac ba \).

Subtracting: \( \log\frac{a + ib}{a - ib} = 2i\tan^{-1}\frac ba \).

Q6. Prove that \( \log(e^{i\alpha} + e^{i\beta}) = \log\left[2\cos\frac{\alpha - \beta}{2}\right] + i\frac{\alpha + \beta}{2} \) (principal value, assuming \( \cos\frac{\alpha - \beta}{2} > 0 \) and \( |\alpha + \beta| < 2\pi \)). (4 marks)

\( e^{i\alpha} + e^{i\beta} = e^{i(\alpha + \beta)/2}\left[e^{i(\alpha - \beta)/2} + e^{-i(\alpha - \beta)/2}\right] = 2\cos\frac{\alpha - \beta}{2}\,e^{i(\alpha + \beta)/2} \).

This is in the form \( re^{i\theta} \) with \( r = 2\cos\frac{\alpha - \beta}{2} \) and \( \theta = \frac{\alpha + \beta}{2} \).

Hence \( \log(e^{i\alpha} + e^{i\beta}) = \log r + i\theta = \log\left[2\cos\frac{\alpha - \beta}{2}\right] + i\frac{\alpha + \beta}{2} \).

Q7. Find the general and principal values of \( \log(-3) \) and \( \log(1 + i\sqrt3) \). (4 marks)

\( -3 = 3e^{i\pi} \): \( \text{Log}(-3) = \log3 + i(2n + 1)\pi \); principal value \( \log3 + i\pi \).

\( 1 + i\sqrt3 = 2e^{i\pi/3} \): \( \text{Log}(1 + i\sqrt3) = \log2 + i\left(2n\pi + \frac\pi3\right) \); principal value \( \log2 + i\frac\pi3 \).

Q8. Prove that \( \cosh^2x - \sinh^2x = 1 \) and \( \cosh2x = 1 + 2\sinh^2x \) from the definitions. (3 marks)

\( \cosh^2x - \sinh^2x = \frac{(e^x + e^{-x})^2 - (e^x - e^{-x})^2}{4} = \frac{4}{4} = 1 \).

\( 1 + 2\sinh^2x = 1 + \frac{e^{2x} - 2 + e^{-2x}}{2} = \frac{e^{2x} + e^{-2x}}{2} = \cosh2x \).

Q9. Separate into real and imaginary parts the principal value of \( (\sqrt i)^{\sqrt i} \). (6 marks)

\( \sqrt i = e^{i\pi/4} = \frac{1 + i}{\sqrt2} \) (principal value), so \( \log\sqrt i = i\frac\pi4 \).

\( \sqrt i\log\sqrt i = \frac{1 + i}{\sqrt2}\cdot\frac{i\pi}{4} = \frac{\pi}{4\sqrt2}(i - 1) = -\frac{\pi}{4\sqrt2} + i\frac{\pi}{4\sqrt2} \).

\( (\sqrt i)^{\sqrt i} = e^{-\pi/(4\sqrt2)}\left[\cos\frac{\pi}{4\sqrt2} + i\sin\frac{\pi}{4\sqrt2}\right] \).

Real part \( e^{-\pi/(4\sqrt2)}\cos\frac{\pi}{4\sqrt2} \), imaginary part \( e^{-\pi/(4\sqrt2)}\sin\frac{\pi}{4\sqrt2} \).

Q10. Prove that \( \tan(x + iy) = \frac{\sin2x + i\sinh2y}{\cos2x + \cosh2y} \). (4 marks)

\( \tan(x + iy) = \frac{2\sin(x + iy)\cos(x - iy)}{2\cos(x + iy)\cos(x - iy)} \).

Numerator: \( \sin2x + \sin2iy = \sin2x + i\sinh2y \) (using 2 sin A cos B = sin(A + B) + sin(A − B)).

Denominator: \( \cos2x + \cos2iy = \cos2x + \cosh2y \) (using 2 cos A cos B = cos(A + B) + cos(A − B)).

Dividing gives the result.

Exam tips

• Memorise the six basic conversions (sin ix, cos ix, tan ix and sinh ix, cosh ix, tanh ix); nearly every question uses them.
• In f(α + iβ) = x + iy problems, write the equations for x and y first and state which identity you use to eliminate α or β.
• When taking a square root such as sinh u = √(cosh²u − 1), mention the sign assumption.
• For logarithm questions, write both the general value (with 2nπ) and the principal value unless only one is asked.
• For complex powers, compute b·log a and split it into A + iB before exponentiating; most errors come from multiplying out this product.
• Check the quadrant of the argument before writing log z.

FAQs

Why are sinh and cosh called hyperbolic functions?

Because the point (cosh t, sinh t) lies on the hyperbola x² − y² = 1, just as (cos t, sin t) lies on the circle x² + y² = 1.

What is the difference between Log z and log z?

Log z denotes the general many-valued logarithm log r + i(θ + 2nπ); log z usually denotes its principal value with n = 0.

Can sin z be greater than 1 for complex z?

Yes. For example sin(iy) = i sinh y and cos(iy) = cosh y, which grows without bound, so circular functions of complex numbers are not bounded by 1.

Do I need to prove the separation formulas in the exam?

Usually a short derivation using the addition formulas and the identities cos iy = cosh y, sin iy = i sinh y is expected, especially for tan(x + iy) and tanh(x + iy).

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Applied Mathematics-I) and were prepared with AI help and checked.