Complex Analysis 12 | Exp, Cos and Sin as Power Series [dark version]

TL;DR

Holomorphic functions given by power series have derivatives that are also holomorphic and represented by power series with the same disk of convergence.

Briefing Cornell Notes

Briefing

Holomorphic power series behave predictably under differentiation: if a function is given by a power series on its disk of convergence, then every derivative exists everywhere on that same disk and is itself represented by a power series with the same radius of convergence. That “derivatives stay holomorphic” principle is the backbone for the examples that follow, and it matters because it lets complex derivatives be computed term-by-term and keeps analytic structure intact.

The lesson starts with the exponential function. Using its standard power series expansion, the radius of convergence comes out infinite, so exp(z) is an entire function (holomorphic on all of ℂ). Differentiating term-by-term produces a new series where the factor k from differentiating z^k cancels the k! in the denominator, leaving (after an index shift) the original series back. The conclusion is that exp(z) is unchanged by differentiation: every derivative of the exponential function equals exp(z). This isn’t just a computational trick; it’s a defining stability property that later simplifies many complex-variable calculations.

Next comes cosine, built from its complex power series. The series contains only even powers of z, and the coefficients alternate signs: z^2 carries a minus sign, z^4 is positive, z^6 is negative, and so on. Rather than differentiate that series directly, the discussion connects cosine to exponentials using the identity i^2 = −1. By substituting i z into the exponential series, the powers of i cycle through four cases (1, i, −1, −i), producing a structured pattern of coefficients. To eliminate the unwanted odd-power contributions involving i, the method introduces a second exponential with −i z. Adding e^{iz} and e^{−iz} cancels the terms with imaginary factors and leaves only the even-power terms—exactly the structure needed for cosine.

The result is the classic representation cos(z) = (1/2) (e^{iz} + e^{−iz}). With this in hand, differentiating becomes straightforward. Applying the chain rule to each exponential term yields a factor of i from e^{iz} and a factor of −i from e^{−iz}; the exponentials themselves remain unchanged under differentiation. The combined effect gives the familiar derivative relationship d/dz [cos(z)] = −sin(z), now expressed through exponentials as well.

The takeaway is practical: exponential, cosine, and sine are all holomorphic, and rewriting trigonometric functions in terms of exponentials turns many algebraic manipulations into simpler exponential-series calculations—especially when powers and derivatives are involved.

Cornell Notes

Holomorphic functions defined by power series keep their analytic structure under differentiation: differentiating does not shrink the disk of convergence, and every derivative is again holomorphic with a power-series representation on the same disk. For exp(z), the power series has infinite radius of convergence, making it entire, and term-by-term differentiation reproduces the same series—so all derivatives of exp(z) equal exp(z). Cos(z) has a power series with only even powers and alternating signs. By using i^2 = −1, cos(z) can be rewritten as cos(z) = (1/2)(e^{iz} + e^{−iz}), which makes differentiation easy and yields d/dz[cos(z)] = −sin(z). This exponential form also simplifies later computations involving sine and cosine.

Why does differentiating a power-series-defined holomorphic function keep it holomorphic on the same disk of convergence?

If f(z) is given by a power series on some disk of convergence, then f′(z) exists everywhere on that disk and is represented by another power series with the same radius of convergence. Since f′ is holomorphic on that disk, the same reasoning applies again to f′′, and so on. The key point is that the power-series structure survives differentiation without reducing the convergence domain.

How does term-by-term differentiation of exp(z) return exp(z) itself?

Starting from exp(z) = Σ_{k=0}^∞ z^k/k!, differentiating gives Σ_{k=1}^∞ k z^{k−1}/k!. The factor k cancels with k! to produce z^{k−1}/(k−1)!. After an index shift (letting m = k−1), the series becomes Σ_{m=0}^∞ z^m/m!, which is exactly exp(z). The constant term disappears in differentiation because k=0 contributes nothing.

What feature of the cosine power series forces only even powers of z to appear?

Cos(z) is defined by a series where the coefficients alternate signs in a way tied to powers of i. In the resulting cosine series, only even powers occur: z^2 has a minus sign, z^4 is positive, z^6 is negative, etc. This even-only structure is what later allows cancellation of odd-power terms when combining e^{iz} and e^{−iz}.

How does adding e^{iz} and e^{−iz} eliminate the unwanted imaginary contributions?

Expanding e^{iz} and e^{−iz} via their power series introduces factors of i^k. Those factors cycle through four values (1, i, −1, −i). When the two exponentials are added, terms with odd k carry opposite imaginary factors and cancel, while even k terms match and remain. The surviving even-power series matches the cosine series up to a factor of 1/2.

Why does differentiating cos(z) written as (1/2)(e^{iz}+e^{−iz}) produce −sin(z)?

Differentiate term-by-term using the chain rule: d/dz[e^{iz}] = i e^{iz} and d/dz[e^{−iz}] = (−i)e^{−iz}. Multiplying by 1/2 gives (i/2)e^{iz} − (i/2)e^{−iz} = (i/2)(e^{iz} − e^{−iz}). That expression is the exponential form of −sin(z), so the derivative is −sin(z).

Review Questions

What changes (or doesn’t change) about the radius of convergence when differentiating a power series?
Show the cancellation step that turns k/k! into 1/(k−1)! when differentiating exp(z).
Using cos(z) = (1/2)(e^{iz}+e^{−iz}), compute d/dz[cos(z)] and identify the result in terms of sine.

Key Points

1
Holomorphic functions given by power series have derivatives that are also holomorphic and represented by power series with the same disk of convergence.
2
exp(z) is entire because its power series has infinite radius of convergence.
3
Term-by-term differentiation of exp(z) reproduces the original series, so every derivative of exp(z) equals exp(z).
4
cos(z) has a power series containing only even powers of z with alternating signs.
5
The identity cos(z) = (1/2)(e^{iz}+e^{−iz}) comes from canceling odd-power terms when adding e^{iz} and e^{−iz}.
6
Writing cosine in exponential form makes differentiation straightforward and yields d/dz[cos(z)] = −sin(z).
7
Exponential, cosine, and sine are all holomorphic, enabling power-series and chain-rule methods throughout complex analysis.

Highlights

Differentiation of a power-series-defined holomorphic function preserves the same radius of convergence, so all derivatives remain holomorphic on that disk.

The exponential function is uniquely stable under differentiation: exp(z) = d/dz[exp(z)], so all derivatives equal exp(z).

Cosine can be rebuilt from exponentials: cos(z) = (1/2)(e^{iz}+e^{−iz}), with odd-power contributions canceling automatically.

Using the exponential form turns the derivative of cosine into a quick chain-rule computation, giving −sin(z).