By Vivek Sharma·MATHEMATICS·Ch. 2 · Complex NumbersFriday, October 23, 2026· 18 min read

Complex Series and Trigonometric Functions

Extend convergence, power series, sine, and cosine to complex inputs, one step at a time.

You now know how to calculate with complex numbers. The next question is what happens when we add infinitely many of them or put a complex number into a familiar function. This lesson uses partial sums and power series from Chapter 1. We will review the essential definitions before extending them.

Before you begin: Complex Numbers: Start with the Basics →

1. What does convergence mean in the plane?

A sequence $z_n$ converges to $L$ when the distance $|z_n-L|$ approaches zero. The points must settle towards a single point in the complex plane. This happens exactly when the real parts approach the real part of $L$ and the imaginary parts approach its imaginary part.

For a complex series, first form the partial sum $S_N=z_1+z_2+\cdots+z_N$ . The series converges if $S_N$ approaches a finite complex number. Equivalently, both component series must converge.

\sum_{n=1}^{\infty}(a_n+ib_n)=\sum_{n=1}^{\infty}a_n+i\sum_{n=1}^{\infty}b_n\quad\text{if both sums converge}

Worked example · Separate the components

Consider $\sum_{n=1}^{\infty}(1+i)/2^n$ .
Its real part is $\sum_{n=1}^{\infty}1/2^n=1$ . Its imaginary coefficient has the same sum.
Both converge, so the complex sum is $1+i$ .

Remember: A complex sum converges when both of its real component sums converge.

2. Absolute convergence gives a useful test

A series is absolutely convergent if the real, nonnegative series $\sum|z_n|$ converges. Absolute convergence guarantees convergence of the complex series. It is a sufficient condition, not a necessary one.

The term test still applies: if $z_n$ does not approach zero, the series diverges. Terms approaching zero do not guarantee convergence; the harmonic series remains a counterexample.

The ratio test uses the limit $L$ below when it exists and the ratios are eventually defined. If $L<1$ , the series converges absolutely. If $L>1$ (including an infinite limit), it diverges. If $L=1$ , use another test.

L=\lim_{n\to\infty}\left|\frac{z_{n+1}}{z_n}\right|

Worked example · Magnitudes remove rotating signs

For $\sum_{n=1}^{\infty}i^n/n^2$ , each $|i^n|=1$ , so the absolute-value series is $\sum1/n^2$ .
That real p-series converges because $p=2>1$ . Therefore the complex series converges absolutely.
We have proved convergence without finding the exact sum. These are different tasks.

Remember: Taking moduli often turns a complex convergence problem into a familiar real one.

3. A geometric ratio can be complex

A geometric series has first term $A$ and common ratio $q$ , which may now be complex. The finite-sum derivation uses only algebra, so it still works. When $|q|<1$ , the modulus of $q^N$ tends to zero, giving the infinite sum.

For a nonzero first term, the series diverges when $|q|\geq1$ : its terms do not approach zero. If $A=0$ , every term is zero.

\sum_{n=0}^{N-1}Aq^n=\frac{A(1-q^N)}{1-q}\quad(q\ne1)

\sum_{n=0}^{\infty}Aq^n=\frac{A}{1-q}\quad(|q|<1)

Worked example · Sum 1 + i/2 + (i/2)² + ⋯

Here $A=1$ , $q=i/2$ , and $|q|=1/2<1$ .
The sum is $1/(1-i/2)$ . Multiply numerator and denominator by $1+i/2$ .
This gives $(1+i/2)/(1+1/4)=4/5+(2/5)i$ .

Remember: Check the modulus of the ratio before summing a complex geometric series.

4. An interval becomes a disc

A complex power series has the form $\sum a_n(z-c)^n$ , where the coefficients and centre may be complex. It converges absolutely inside a radius $R$ about $c$ and diverges outside. Boundary points must be checked separately.

For real inputs, $|x-c|<R$ describes an interval. In the complex plane, $|z-c|<R$ describes the inside of a circle, called an open disc. A power series always converges at its centre. A radius can also be zero or infinite.

\sum_{n=0}^{\infty}\frac{(z-1)^n}{2^n}=\frac{1}{1-(z-1)/2}\qquad(|z-1|<2)

Worked example · Check a complex input

The centre of this series is $1$ and its radius is $2$ , since its ratio is $(z-1)/2$ .
At $z=1+i$ , $|z-1|=|i|=1<2$ , so it converges. The sum is $1/(1-i/2)=4/5+(2/5)i$ .
On the boundary $|z-1|=2$ , the ratio has modulus one. In this particular geometric example, every boundary point diverges because the terms have modulus one. Other power series can have different boundary behaviour.

Remember: The same radius idea works in two dimensions: replace an interval by a disc.

5. Define the complex exponential by a series

For a complex input $z$ , define $e^z$ by the same power series used for real inputs. Here $n!$ means the factorial: $n!=1\times2\times\cdots\times n$ , with $0!=1$ .

For fixed $z\ne0$ , the ratio of successive term magnitudes is $|z|/(n+1)$ , which tends to zero. At $z=0$ , the series equals $1$ . Thus the exponential series converges absolutely for every complex $z$ .

A finite truncation is an approximation, not an exact equality. The usual exponential law $e^{u+v}=e^u e^v$ remains valid for complex numbers; it can be derived by multiplying the absolutely convergent series and collecting equal powers.

e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots

Worked example · Approximate e^(0.1i)

Keep terms through $z^3$ with $z=0.1i$ .
Since $(0.1i)^2=-0.01$ and $(0.1i)^3=-0.001i$ , the approximation is $1+0.1i-0.01/2-0.001i/6$ .
So $e^{0.1i}\approx0.995+0.0998333i$ . The omitted tail has modulus at most $e^{|z|}|z|^4/4!<0.00000461$ here. This bound follows by bounding the magnitudes of the remaining exponential-series terms.

Remember: The exponential series works for every complex input, and a short polynomial can approximate it.

6. Sine and cosine also accept complex inputs

Define complex sine and cosine by the power series below. Both converge absolutely for every complex $z$ , using the ratio test on successive nonzero series terms. For real inputs, these agree with the familiar trigonometric functions, with angles measured in radians.

To evaluate an imaginary input, substitute it into the series and simplify powers of $i$ . The alternating signs can disappear. For example, $(iy)^{2n}=(-1)^n y^{2n}$ for real $y$ .

\cos z=\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n}}{(2n)!}=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots

\sin z=\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n+1}}{(2n+1)!}=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots

Worked example · Substitute z = i

For cosine, $\cos i=1-i^2/2!+i^4/4!-\cdots=1+1/2!+1/4!+\cdots$ .
For sine, $\sin i=i-i^3/3!+i^5/5!-\cdots=i(1+1/3!+1/5!+\cdots)$ .
Thus $\cos i$ is real and greater than $1$ , while $\sin i$ is purely imaginary. The familiar real-input bound $|\cos x|\leq1$ does not extend to all complex inputs.

Remember: Power series extend sine and cosine consistently beyond the real line.

7. A useful bridge to hyperbolic functions

Two related functions appear naturally when the input is imaginary. For real $y$ , define hyperbolic cosine and hyperbolic sine using the real exponentials below. They are written cosh and sinh.

Expanding $e^y$ and $e^{-y}$ shows that cosh contains the even powers and sinh the odd powers. Comparing with the sine and cosine series gives $\cos(iy)=\cosh y$ and $\sin(iy)=i\sinh y$ .

The identity $\cos^2z+\sin^2z=1$ still holds for complex inputs. At $z=iy$ , it becomes $\cosh^2y-\sinh^2y=1$ , because $i^2=-1$ . These functions later appear in differential equations and models of hanging cables.

\cosh y=\frac{e^y+e^{-y}}2,\qquad\sinh y=\frac{e^y-e^{-y}}2

\cos(iy)=\cosh y,\qquad\sin(iy)=i\sinh y

Remember: Imaginary inputs connect trigonometric functions to real exponential growth and decay.