Hyperbolic and Trigonometric Functions in the Complex Domain

In the complex domain, sine and cosine are defined by exponentials, sin z = (e^iz − e^−iz)/(2i) and cos z = (e^iz + e^−iz)/2, and they are related to the hyperbolic functions by cos(iz) = cosh z and sin(iz) = i sinh z. In brief: sin and cos keep all their identities, such as sin²z + cos²z = 1, but they are no longer bounded: cos(2i) = cosh 2 ≈ 3.762. For z = x + iy, sin z = sin x cosh y + i cos x sinh y.

Euler’s formula connects exponentials with trigonometry, and this connection is the reason why the hyperbolic functions appear as the “trigonometric functions of an imaginary angle”. The extension of sine and cosine to complex arguments solves equations such as cos z = 2 and explains the formulas for hyperbolic functions. This guide defines the functions, derives the relations and addition formulas, treats the real and imaginary parts, solves equations and gives a table of values of sinh and cosh.

How are the functions defined?

From Euler’s formula, e^iz = cos z + i sin z for every complex z. Adding and subtracting e^iz and e^−iz gives

• cos z = (e^iz + e^−iz)/2,
• sin z = (e^iz − e^−iz)/(2i).

The hyperbolic functions are

• cosh z = (e^z + e^−z)/2,
• sinh z = (e^z − e^−z)/2,
• tanh z = sinh z/cosh z, and similarly tan z = sin z/cos z.

For real x, cosh x ≥ 1 is the shape of a hanging chain (the catenary), and sinh x is an odd function growing like e^x/2. The identity cosh²z − sinh²z = 1 shows that (cosh t, sinh t) lies on the hyperbola x² − y² = 1, just as (cos t, sin t) lies on the circle.

How are trigonometric and hyperbolic functions related?

Substituting iz into the definitions gives:

• cos(iz) = cosh z and cosh(iz) = cos z,
• sin(iz) = i sinh z and sinh(iz) = i sin z,
• tan(iz) = i tanh z.

Proof of cos(iz) = cosh z: cos(iz) = (e^i·iz + e^−i·iz)/2 = (e^−z + e^z)/2 = cosh z ✓.

The identities of trigonometry hold for complex arguments, since they follow from the exponential law: sin²z + cos²z = 1, sin(z + w) = sin z cos w + cos z sin w, cos(z + w) = cos z cos w − sin z sin w. Replacing z by iz in sin²z + cos²z = 1 gives −sinh²z + cosh²z = 1. In this way the formulas for hyperbolic functions follow from trigonometric formulas by “Osborn’s rule”: replace each product of two sines by minus the product of two sinh. Both pairs are periodic in the complex domain: sin and cos have the real period 2π, and sinh and cosh have the imaginary period 2πi.

What are the real and imaginary parts of sin z and cos z?

For z = x + iy:

• sin(x + iy) = sin x cosh y + i cos x sinh y,
• cos(x + iy) = cos x cosh y − i sin x sinh y,
• |sin z|² = sin²x + sinh²y and |cos z|² = cos²x + sinh²y.

These come from the addition formula with sin(iy) = i sinh y and cos(iy) = cosh y. Since sinh y grows without bound, sin z and cos z are unbounded in the complex plane, though bounded by 1 on the real axis. The only zeros of sin z are the real numbers kπ, since |sin z|² = 0 requires sin x = 0 and sinh y = 0, so y = 0.

Example 1

sin(i) = i sinh 1 = i · 1.1752 = 1.1752i. And cos(2i) = cosh 2 = 3.7622. The magnitude |sin(i)| ≈ 1.175 > 1.

Example 2

sin(1 + i) = sin 1 cosh 1 + i cos 1 sinh 1 = 0.8415 · 1.5431 + i · 0.5403 · 1.1752 = 1.2985 + 0.6350i. Check the modulus: |sin(1 + i)|² = sin²1 + sinh²1 = 0.7081 + 1.3811 = 2.0892, and 1.2985² + 0.6350² = 1.6861 + 0.4032 = 2.0893 ✓.

Example 3

cos(π + i) = cos π cosh 1 − i sin π sinh 1 = −1.5431 (real, since sin π = 0).

How do you solve equations?

To solve sin z = w or cos z = w, write the equation in terms of u = e^iz and solve a quadratic equation, then take the logarithm.

Example 4: cos z = 2

(u + 1/u)/2 = 2, so u² − 4u + 1 = 0, giving u = 2 ± √3. Then e^iz = 2 ± √3, so iz = ln(2 ± √3) + 2kπi, and z = 2kπ − i ln(2 ± √3). Since ln(2 − √3) = −ln(2 + √3), the solutions are z = 2kπ ± i ln(2 + √3) ≈ 2kπ ± 1.3170i. Check with k = 0: cos(1.317i) = cosh 1.317 = 2.000 ✓.

Example 5: sin z = 2

(u − 1/u)/(2i) = 2, so u² − 4iu − 1 = 0, giving u = 2i ± √(−4 + 1) = i(2 ± √3). Then iz = ln(2 ± √3) + iπ/2 + 2kπi, so z = π/2 + 2kπ − i ln(2 ± √3), that is, z = π/2 + 2kπ ± i ln(2 + √3). Check for k = 0: sin(π/2 + 1.317i) = cos(1.317i) = cosh 1.317 = 2 ✓.

Example 6: the hyperbolic equation sinh z = 0

(e^z − e^−z)/2 = 0 gives e^2z = 1, so 2z = 2kπi and z = kπi. The zeros of sinh are on the imaginary axis, consistent with sinh(iz) = i sin z.

Example 7: a series check

The power series sin z = z − z³/3! + z⁵/5! − … converges for every complex z. For z = i: i − i³/6 + i⁵/120 − … = i(1 + 1/6 + 1/120 + …) = i sinh 1 ✓.

Table of hyperbolic values

Where do these functions come from?

Hyperbolic functions were introduced by Vincenzo Riccati in the 1760s and developed by Johann Heinrich Lambert, who also studied their relation to circular functions. The link with complex numbers, through Euler’s formula, was clear to Euler. The extension of trigonometric functions to complex arguments belongs to the work of Euler and Cauchy and leads to the general theory of analytic functions.

Where are they used today?

The catenary, the shape of a hanging chain or cable, is y = a cosh(x/a). The Gateway Arch in St. Louis approximates an inverted catenary. In relativity, rapidity is a hyperbolic angle: velocities add by the hyperbolic tangent addition formula. In engineering, solutions of the wave and heat equations involve sin, cos, sinh and cosh, and the transmission line equations use complex hyperbolic functions.

Which mistakes are common?

Assuming |sin z| ≤ 1

This holds only for real z. In the complex plane, sin z and cos z are unbounded.

Mixing up sign changes

sin(iz) = i sinh z, with a factor i. cos(iz) = cosh z, with no factor.

Forgetting the multi-valued solutions

cos z = 2 has infinitely many solutions z = 2kπ ± 1.317i.

Using the real identities with a sign error

cosh²z − sinh²z = 1 has a minus sign, unlike sin²z + cos²z = 1.