Where do sine and cosine even come from?
In the previous guide you extended the exponential to the whole plane and watched e^z gain a 2 pi i period. Now we ride that work straight into trigonometry. The bridge is Euler's formula, which you already trust for real angles: e^(i t) = cos t + i sin t. Replace t by -t and use the fact that cosine is even and sine is odd, and you also get e^(-i t) = cos t - i sin t. Two equations, two unknowns — solve them for cos t and sin t.
Adding the two identities cancels the sines and gives cos t = (e^(i t) + e^(-i t))/2; subtracting them cancels the cosines and gives sin t = (e^(i t) - e^(-i t))/(2 i). These hold for every real t. Now comes the one move that creates the whole theory: just let the angle become a complex number z. We define the complex sine and cosine by cos z = (e^(i z) + e^(-i z))/2 and sin z = (e^(i z) - e^(-i z))/(2 i), with z = x + i y now free to wander anywhere in the plane.
This is not a guess we hope behaves well; it is the only honest choice. Because e^z is the unique entire function agreeing with the real exponential on the axis, the cos z and sin z built from it are the unique entire functions agreeing with ordinary cosine and sine on the real line. The identity theorem from the holomorphic rung guarantees there is no rival — any other entire function matching them on the real axis would have to equal them everywhere.
What survives, and what changes
The comforting news first: nearly every algebraic identity you memorized survives the trip to the plane unchanged. The Pythagorean identity cos^2 z + sin^2 z = 1 still holds for all complex z, the angle-addition formulas still work, and the derivatives are exactly what you expect — the derivative of sin z is cos z, and the derivative of cos z is -sin z. None of this is a coincidence: any polynomial relation between cosine and sine that is true along the whole real axis must, by the identity theorem again, be true everywhere it makes sense.
The periodicity survives too, and it comes straight from the exponential's. Since e^z repeats every time z increases by 2 pi i, the functions e^(i z) and e^(-i z) repeat every time z increases by 2 pi. So cos z and sin z keep their real period 2 pi exactly — and that is the only period. There is no extra imaginary period; pushing z off the real axis does not bring you back to where you started. Keep that firmly in mind, because the behavior in the imaginary direction is about to break a rule you have trusted since high school.
The cage opens: unboundedness
Here is the headline of this whole guide. On the real line sin t and cos t are forever boxed inside [-1, 1]. In the plane that box is gone: the complex sine and cosine are unbounded. To see it, do not look along the real axis at all — climb the imaginary axis. Put z = i y, a purely imaginary input, and feed it into the definition. Then e^(i z) = e^(-y) and e^(-i z) = e^(y), and the formulas collapse into something startling.
Set z = i y (pure imaginary). Then:
cos(i y) = ( e^(-y) + e^(y) ) / 2 = cosh y
sin(i y) = ( e^(-y) - e^(y) ) / (2 i) = i sinh y
As y -> +infinity, cosh y and sinh y both -> +infinity.
example: |cos(10 i)| = cosh 10 is about 11013,
nowhere near the old bound of 1.Read that slowly. Along the imaginary axis, cos z is exactly the real hyperbolic cosine cosh y, which sails off to infinity as y grows. So |cos z| is as large as you like once you go high enough — at z = 10 i it is already over eleven thousand. The same exponential growth happens for sin z. The familiar wave that never rose above 1 was a special, flat slice through a function that, seen in full, grows without any ceiling at all the moment you step off the real line.
Why this had to happen: Liouville keeps the books
The unboundedness is not an accident or a quirk of the formula. It is forced. Recall Liouville's theorem from the holomorphic rung: a function that is entire (holomorphic on the whole plane) AND bounded must be constant. Now cos z and sin z are entire — built from e^z, which is entire — and they are plainly not constant. So they simply cannot be bounded. If they stayed inside [-1, 1] everywhere, Liouville would crush them down to constants, which they are not. The cage had to open.
This is a beautiful example of the rigidity complex analysis keeps delivering. In real analysis you can easily build a smooth, non-constant function that stays bounded forever — sin t itself does it. But complex differentiability is far stronger than real differentiability, and that extra strength has a price: it leaves no room for a bounded, non-constant entire function. The very thing that makes cos z so well-behaved (being holomorphic everywhere) is exactly what forces it to grow.
- Recall the definition: cos z = (e^(i z) + e^(-i z)) / 2, valid for every complex z.
- Note that e^z is entire, so e^(i z) and e^(-i z) are entire, and hence cos z is entire.
- Suppose, for contradiction, that |cos z| stayed below some fixed bound M for all z.
- Then Liouville's theorem forces cos z to be constant — but cos 0 = 1 and cos pi = -1, a contradiction. So no such bound exists: cos z is unbounded.
Sine, cosine, and the hyperbolic twins are one family
The computation cos(i y) = cosh y was not a stray fact — it points to a deep merger. The complex hyperbolic functions cosh z = (e^z + e^(-z))/2 and sinh z = (e^z - e^(-z))/2 are built from the same exponential, so they are entire too, and over the complex numbers they are nothing more than the trig functions rotated. The clean statements are cosh z = cos(i z) and sinh z = -i sin(i z). On the real line cos and cosh look like utterly different beasts — one a wave, one a hanging chain — but in the plane they are the same function viewed from perpendicular directions.
Splitting into real and imaginary parts makes the whole family concrete. Using the addition formula, sin(x + i y) = sin x cosh y + i cos x sinh y, and cos(x + i y) = cos x cosh y - i sin x sinh y. Read these as the f = u + i v decomposition from the functions rung: the real part oscillates with x (the old wave) while the imaginary part carries the runaway sinh y growth. You can even check the Cauchy-Riemann equations on these u and v and watch them hold — a hands-on confirmation that sin z really is holomorphic.
One more honest caveat before you go. Because sine and cosine now hit every value and grow without bound, the equation sin z = w can be solved for any complex w — including w with |w| far above 1. Solving it leads you to inverse functions like arcsin built from a square root and a logarithm, and those will turn out to be multivalued, with the same branch-point trouble the next two guides explore for the logarithm. The pleasant single-valued world of sin and cos is real, but its inverses already smell of the multivaluedness coming next.