Power series for real numbers was covered in semester 1. This note explains the complex numbers version.
A complex power series centered at α \alpha α is a series of the form
∑ n = 0 ∞ c n ( z − α ) n \sum_{n=0}^{\infty} c_n (z-\alpha)^n n = 0 ∑ ∞ c n ( z − α ) n
Here:
α \alpha α : center, a fixed complex number
c n c_n c n : coefficients, fixed complex numbers
z z z : a complex variable
Radius of Convergence
The radius of convergence ρ \rho ρ is defined as in the real case . For complex z z z , the interval of convergence becomes a disk.
There exists ρ \rho ρ with 0 ≤ ρ ≤ ∞ 0 \le \rho \le \infty 0 ≤ ρ ≤ ∞ such that the series
converges for ∣ z − α ∣ < ρ |z-\alpha| < \rho ∣ z − α ∣ < ρ
diverges for ∣ z − α ∣ > ρ |z-\alpha| > \rho ∣ z − α ∣ > ρ
Ignoring boundary points, the convergence region is 1 of the following.
Single point
The point α \alpha α , when ρ = 0 \rho = 0 ρ = 0 .
Disk
D ρ ( α ) = { z ∈ C : ∣ z − α ∣ < ρ } D_\rho(\alpha) = \{z \in \mathbb{C} : |z-\alpha| < \rho\} D ρ ( α ) = { z ∈ C : ∣ z − α ∣ < ρ } , when 0 < ρ < ∞ 0 < \rho < \infty 0 < ρ < ∞ .
Entire complex plane
When ρ = ∞ \rho = \infty ρ = ∞ .
Convergence on the boundary circle ∣ z − α ∣ = ρ |z-\alpha| = \rho ∣ z − α ∣ = ρ , when 0 < ρ < ∞ 0 < \rho < \infty 0 < ρ < ∞ , must be checked separately at each point.
Examples:
∑ n = 0 ∞ ( n + 1 ) ( z − 2 ) n 3 n \sum_{n=0}^{\infty} \frac{(n+1)(z-2)^n}{3^n} ∑ n = 0 ∞ 3 n ( n + 1 ) ( z − 2 ) n has ρ = 3 \rho = 3 ρ = 3 and disk of convergence D 3 ( 2 ) D_3(2) D 3 ( 2 )
Algebra of Power Series
Let f f f and g g g have power series representations
f ( z ) = ∑ n = 0 ∞ a n ( z − α ) n , z ∈ D r 1 ( α ) f(z) = \sum_{n=0}^{\infty} a_n (z-\alpha)^n, \quad z \in D_{r_1}(\alpha) f ( z ) = n = 0 ∑ ∞ a n ( z − α ) n , z ∈ D r 1 ( α )
g ( z ) = ∑ n = 0 ∞ b n ( z − α ) n , z ∈ D r 2 ( α ) g(z) = \sum_{n=0}^{\infty} b_n (z-\alpha)^n, \quad z \in D_{r_2}(\alpha) g ( z ) = n = 0 ∑ ∞ b n ( z − α ) n , z ∈ D r 2 ( α )
Let r = min { r 1 , r 2 } r = \min\{r_1, r_2\} r = min { r 1 , r 2 } and β ∈ C \beta \in \mathbb{C} β ∈ C . Then for z ∈ D r ( α ) z \in D_r(\alpha) z ∈ D r ( α ) :
Scalar multiple
β f ( z ) = ∑ n = 0 ∞ β a n ( z − α ) n \beta f(z) = \sum_{n=0}^{\infty} \beta a_n (z-\alpha)^n β f ( z ) = ∑ n = 0 ∞ β a n ( z − α ) n
Sum
f ( z ) + g ( z ) = ∑ n = 0 ∞ ( a n + b n ) ( z − α ) n f(z) + g(z) = \sum_{n=0}^{\infty} (a_n + b_n)(z-\alpha)^n f ( z ) + g ( z ) = ∑ n = 0 ∞ ( a n + b n ) ( z − α ) n
Product
f ( z ) g ( z ) = ∑ n = 0 ∞ c n ( z − α ) n f(z)g(z) = \sum_{n=0}^{\infty} c_n (z-\alpha)^n f ( z ) g ( z ) = ∑ n = 0 ∞ c n ( z − α ) n
The coefficients of the product are
c n = ∑ k = 0 n a k b n − k c_n = \sum_{k=0}^{n} a_k b_{n-k} c n = k = 0 ∑ n a k b n − k
This is the Cauchy product of the series.
Examples:
The first 4 non-zero terms of the Maclaurin series of e z sin z e^z \sin z e z sin z are z + z 2 + z 3 3 − z 5 30 z + z^2 + \frac{z^3}{3} - \frac{z^5}{30} z + z 2 + 3 z 3 − 30 z 5
Worked Examples
Radius and Boundary
Find the disk of convergence of ∑ n = 0 ∞ ( n + 1 ) ( z − 2 ) n 3 n \sum_{n=0}^{\infty} \frac{(n+1)(z-2)^n}{3^n} ∑ n = 0 ∞ 3 n ( n + 1 ) ( z − 2 ) n .
Apply the ratio test .
lim n → ∞ ∣ u n + 1 u n ∣ = lim n → ∞ ( n + 2 ) ∣ z − 2 ∣ 3 ( n + 1 ) = ∣ z − 2 ∣ 3 \lim_{n\to \infty} \left|\frac{u_{n+1}}{u_n}\right| =
\lim_{n\to \infty} \frac{(n+2)\,\lvert z-2 \rvert}{3(n+1)} =
\frac{|z-2|}{3} n → ∞ lim u n u n + 1 = n → ∞ lim 3 ( n + 1 ) ( n + 2 ) ∣ z − 2 ∣ = 3 ∣ z − 2∣
The series converges for ∣ z − 2 ∣ 3 < 1 \frac{|z-2|}{3} < 1 3 ∣ z − 2∣ < 1 , so ρ = 3 \rho = 3 ρ = 3 and the disk is D 3 ( 2 ) D_3(2) D 3 ( 2 ) .
The ratio test is inconclusive on the boundary ∣ z − 2 ∣ = 3 |z-2| = 3 ∣ z − 2∣ = 3 , since the limit equals 1 1 1 . Check the terms directly.
The n n n -th term is u n = ( n + 1 ) ( z − 2 ) n 3 n u_n = \frac{(n+1)(z-2)^n}{3^n} u n = 3 n ( n + 1 ) ( z − 2 ) n . Taking the modulus gives
∣ u n ∣ = ( n + 1 ) ∣ z − 2 ∣ n 3 n = ( n + 1 ) ( ∣ z − 2 ∣ 3 ) n |u_n| = (n+1)\frac{|z-2|^n}{3^n} = (n+1)\left(\frac{|z-2|}{3}\right)^n ∣ u n ∣ = ( n + 1 ) 3 n ∣ z − 2 ∣ n = ( n + 1 ) ( 3 ∣ z − 2∣ ) n
On the boundary ∣ z − 2 ∣ = 3 |z-2| = 3 ∣ z − 2∣ = 3 , so ∣ z − 2 ∣ 3 = 1 \frac{|z-2|}{3} = 1 3 ∣ z − 2∣ = 1 and
∣ u n ∣ = n + 1 → ∞ |u_n| = n+1 \to \infty ∣ u n ∣ = n + 1 → ∞
A convergent series needs u n → 0 u_n \to 0 u n → 0 . Here ∣ u n ∣ |u_n| ∣ u n ∣ grows without bound, so the series diverges at every boundary point, regardless of which point on the circle z z z is.
All 3 Regions
∑ n ! z n \sum n!\, z^n ∑ n ! z n
The ratio is ( n + 1 ) ∣ z ∣ (n+1)|z| ( n + 1 ) ∣ z ∣ , which tends to ∞ \infty ∞ for every z ≠ 0 z \neq 0 z = 0 . So ρ = 0 \rho = 0 ρ = 0 and the region is the point 0 0 0 .
∑ z n \sum z^n ∑ z n
The ratio is ∣ z ∣ |z| ∣ z ∣ , so ρ = 1 \rho = 1 ρ = 1 and the region is D 1 ( 0 ) D_1(0) D 1 ( 0 ) .
∑ z n n ! \sum \frac{z^n}{n!} ∑ n ! z n
The ratio is ∣ z ∣ n + 1 → 0 \frac{|z|}{n+1} \to 0 n + 1 ∣ z ∣ → 0 for every z z z . So ρ = ∞ \rho = \infty ρ = ∞ and the region is the entire plane.
Sum of 2 Series
Let f ( z ) = 1 1 − z = ∑ z n f(z) = \frac{1}{1-z} = \sum z^n f ( z ) = 1 − z 1 = ∑ z n and g ( z ) = 1 1 + z = ∑ ( − 1 ) n z n g(z) = \frac{1}{1+z} = \sum (-1)^n z^n g ( z ) = 1 + z 1 = ∑ ( − 1 ) n z n , both valid in D 1 ( 0 ) D_1(0) D 1 ( 0 ) .
f ( z ) + g ( z ) = ∑ n = 0 ∞ ( 1 + ( − 1 ) n ) z n = 2 ∑ n = 0 ∞ z 2 n = 2 1 − z 2 f(z) + g(z) = \sum_{n=0}^{\infty} \left(1 + (-1)^n\right) z^n = 2\sum_{n=0}^{\infty} z^{2n} = \frac{2}{1-z^2} f ( z ) + g ( z ) = n = 0 ∑ ∞ ( 1 + ( − 1 ) n ) z n = 2 n = 0 ∑ ∞ z 2 n = 1 − z 2 2
The odd coefficients cancel and the even coefficients equal 2 2 2 . The series is valid in D 1 ( 0 ) D_1(0) D 1 ( 0 ) .
Cauchy Product of Exponential and Sine
Find the first 4 non-zero terms of the Maclaurin series of e z sin z e^z \sin z e z sin z .
The coefficients are
a n = 1 n ! , b 1 = 1 , b 3 = − 1 6 , b 5 = 1 120 , b 0 = b 2 = b 4 = 0 a_n = \frac{1}{n!}, \quad b_1 = 1, \; b_3 = -\frac{1}{6}, \; b_5 = \frac{1}{120}, \quad b_0 = b_2 = b_4 = 0 a n = n ! 1 , b 1 = 1 , b 3 = − 6 1 , b 5 = 120 1 , b 0 = b 2 = b 4 = 0
Apply c n = ∑ k = 0 n a k b n − k c_n = \sum_{k=0}^{n} a_k b_{n-k} c n = ∑ k = 0 n a k b n − k .
c 0 = 0 , c 1 = a 0 b 1 = 1 , c 2 = a 1 b 1 = 1 c_0 = 0, \quad c_1 = a_0 b_1 = 1, \quad c_2 = a_1 b_1 = 1 c 0 = 0 , c 1 = a 0 b 1 = 1 , c 2 = a 1 b 1 = 1
c 3 = a 2 b 1 + a 0 b 3 = 1 2 − 1 6 = 1 3 c_3 = a_2 b_1 + a_0 b_3 = \frac{1}{2} - \frac{1}{6} = \frac{1}{3} c 3 = a 2 b 1 + a 0 b 3 = 2 1 − 6 1 = 3 1
c 4 = a 3 b 1 + a 1 b 3 = 1 6 − 1 6 = 0 c_4 = a_3 b_1 + a_1 b_3 = \frac{1}{6} - \frac{1}{6} = 0 c 4 = a 3 b 1 + a 1 b 3 = 6 1 − 6 1 = 0
c 5 = a 4 b 1 + a 2 b 3 + a 0 b 5 = 1 24 − 1 12 + 1 120 = − 1 30 c_5 = a_4 b_1 + a_2 b_3 + a_0 b_5 = \frac{1}{24} - \frac{1}{12} + \frac{1}{120} = -\frac{1}{30} c 5 = a 4 b 1 + a 2 b 3 + a 0 b 5 = 24 1 − 12 1 + 120 1 = − 30 1
e z sin z = z + z 2 + z 3 3 − z 5 30 + ⋯ e^z \sin z = z + z^2 + \frac{z^3}{3} - \frac{z^5}{30} + \cdots e z sin z = z + z 2 + 3 z 3 − 30 z 5 + ⋯
Cauchy Product with Geometric Series
Find the first 5 non-zero terms of the Maclaurin series of e z 1 − z \frac{e^z}{1-z} 1 − z e z .
Here a n = 1 n ! a_n = \frac{1}{n!} a n = n ! 1 and b n = 1 b_n = 1 b n = 1 , so c n = ∑ k = 0 n 1 k ! c_n = \sum_{k=0}^{n} \frac{1}{k!} c n = ∑ k = 0 n k ! 1 .
c 0 = 1 , c 1 = 2 , c 2 = 5 2 , c 3 = 8 3 , c 4 = 65 24 c_0 = 1, \quad c_1 = 2, \quad c_2 = \frac{5}{2}, \quad c_3 = \frac{8}{3}, \quad c_4 = \frac{65}{24} c 0 = 1 , c 1 = 2 , c 2 = 2 5 , c 3 = 3 8 , c 4 = 24 65
e z 1 − z = 1 + 2 z + 5 2 z 2 + 8 3 z 3 + 65 24 z 4 + ⋯ \frac{e^z}{1-z} = 1 + 2z + \frac{5}{2}z^2 + \frac{8}{3}z^3 + \frac{65}{24}z^4 + \cdots 1 − z e z = 1 + 2 z + 2 5 z 2 + 3 8 z 3 + 24 65 z 4 + ⋯
The series is valid in D 1 ( 0 ) D_1(0) D 1 ( 0 ) .