Euler's formula gives $\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$ for real $\theta$. Using the complex exponential, the same formulas define sine and cosine for every complex $z$. On the imaginary axis they turn into the hyperbolic functions, and that is where the surprises live.
Key idea
$$\cos z = \frac{e^{iz} + e^{-iz}}{2}, \qquad \sin z = \frac{e^{iz} - e^{-iz}}{2i}.$$
Both are entire, with $(\sin z)' = \cos z$ and $(\cos z)' = -\sin z$; the usual identities ($\sin^2 z + \cos^2 z = 1$, the sum formulas) still hold. On the imaginary axis, $\cos(iy) = \cosh y$ and $\sin(iy) = i\sinh y$, and in general
$$\sin(x + iy) = \sin x\cosh y + i\cos x\sinh y, \qquad |\sin z|^2 = \sin^2 x + \sinh^2 y.$$
So $\sin z$ and $\cos z$ are unbounded: they take every complex value, including $2$.
Why it works
Put $z = iy$: $\cos(iy) = \frac{e^{-y} + e^{y}}{2} = \cosh y$ and $\sin(iy) = \frac{e^{-y} - e^{y}}{2i} = i\,\frac{e^y - e^{-y}}{2} = i\sinh y$. The sum formula $\sin(x + iy) = \sin x\cos(iy) + \cos x\sin(iy)$ then gives the rectangular form. For the modulus, $|\sin z|^2 = \sin^2 x\cosh^2 y + \cos^2 x\sinh^2 y$; replace $\cosh^2 y$ by $1 + \sinh^2 y$ and use $\sin^2 x + \cos^2 x = 1$.
Worked examples
Worked example: On the imaginary axis
Compute $\cos(i\ln 2)$ and $\sin(i\ln 2)$.
- $\cos(i\ln 2) = \cosh(\ln 2) = \frac{e^{\ln 2} + e^{-\ln 2}}{2} = \frac{2 + \frac{1}{2}}{2} = \frac{5}{4}$.
- $\sin(i\ln 2) = i\sinh(\ln 2) = i\cdot\frac{2 - \frac{1}{2}}{2} = \frac{3}{4}i$.
- Check: $\sin^2 + \cos^2 = -\frac{9}{16} + \frac{25}{16} = 1$.
Answer: $\frac{5}{4}$ and $\frac{3}{4}i$
Worked example: A general point
Write $\sin\left(\frac{\pi}{6} + i\ln 3\right)$ as $a + bi$.
- $\cosh(\ln 3) = \frac{3 + \frac{1}{3}}{2} = \frac{5}{3}$ and $\sinh(\ln 3) = \frac{3 - \frac{1}{3}}{2} = \frac{4}{3}$.
- $\sin\left(\frac{\pi}{6} + i\ln 3\right) = \sin\frac{\pi}{6}\cdot\frac{5}{3} + i\cos\frac{\pi}{6}\cdot\frac{4}{3} = \frac{5}{6} + \frac{2\sqrt{3}}{3}i$.
Answer: $\frac{5}{6} + \frac{2\sqrt{3}}{3}i$
Worked example: Solving $cos z = rac{5}{4}$
Find all solutions of $\cos z = \frac{5}{4}$.
- Put $t = e^{iz}$: $\frac{t + t^{-1}}{2} = \frac{5}{4}$, so $2t^2 - 5t + 2 = 0$ and $t = 2$ or $t = \frac{1}{2}$.
- $e^{iz} = 2$ gives $iz = \ln 2 + 2\pi ik$, so $z = 2\pi k - i\ln 2$; $e^{iz} = \frac{1}{2}$ gives $z = 2\pi k + i\ln 2$.
- All solutions: $z = 2\pi k \pm i\ln 2$. No real solution exists, as expected.
Answer: $z = 2\pi k \pm i\ln 2$
| Still true for complex $z$ | No longer true |
|---|---|
| $\sin^2 z + \cos^2 z = 1$ | $|\sin z| \le 1$ and $|\cos z| \le 1$ |
| the sum and double-angle formulas | $\sin z$ and $\cos z$ are real |
| $\sin z = 0$ only at $z = k\pi$ | $\cos z = 3$ has no solution |
| period $2\pi$ | bounded on the plane |
The hyperbolic functions extend the same way, $\cosh z = \frac{e^z + e^{-z}}{2}$ and $\sinh z = \frac{e^z - e^{-z}}{2}$, and the two families are the same functions turned a quarter turn: $\cos(iz) = \cosh z$ and $\sin(iz) = i\sinh z$ for every complex $z$. So $\cosh$ is periodic with period $2\pi i$, and $\cosh z = 0$ at $z = \frac{\pi i}{2} + k\pi i$.
Because $|\sin z|$ grows like $\frac{1}{2}e^{|\operatorname{Im} z|}$, any estimate that uses $|\sin z| \le 1$ off the real axis is wrong, a common source of errors when bounding contour integrals.
Common mistake
Carrying $|\sin z| \le 1$ over from the real line. The bound comes from the unit circle, which only describes real angles. Off the real axis, $|\sin(x + iy)|^2 = \sin^2 x + \sinh^2 y$, and $\sinh y$ is unbounded: $|\sin(10i)| = \sinh 10 \approx 11{,}013$. Check any claim of boundedness on the imaginary axis first. (Liouville's theorem, later, says a bounded entire function must be constant, so $\sin z$ could not have been bounded.)