Complex Power Series: an Example the Complex Logarithm

Home , Convergent series, Power series, Radius of convergence, Uniform convergence

Introduction Convergence of sequences Convergence of series Sequences of functions Power series The Logarithm

Complex power series: an example The complex logarithm

K. P. Hart

Faculty EEMCS TU Delft

Delft, 29 januari, 2007

K. P. Hart Complex power series: an example Introduction Convergence of sequences Convergence of series Sequences of functions Power series The Logarithm Purpose of this lecture

Recall notions about convergence of real sequences and series Introduce these notions for complex sequences and series Illustrate these using the Taylor series of Log(1 + z)

A readable version of these slides can be found via

http://fa.its.tudelft.nl/~hart

K. P. Hart Complex power series: an example Introduction Convergence of sequences Deﬁnition Convergence of series Examples Sequences of functions Complex sequences Power series The Logarithm The deﬁnition

Deﬁnition

The sequence {xn} converges to the real number x, in symbols,

lim xn = x n→∞ means: for every positive there is a natural number N such that for all n > N one has |xn − x| < .

The following should be well-known: 1 limn→∞ n = 0 1 more generally limn→∞ np = 0 when p > 0. n limn→∞ x = 0 if |x| < 1

√ n We show limn→∞ n = 1. √ √ n n Note n > 1, so an = n − 1 is positive.

Apply the binomial formula:

n X n 1 n = (1 + a )n = ak = 1 + na + n(n − 1)a2 + ··· n k n n 2 n k=0 we drop all terms but the second . . .

1 2 2 2 . . . and we ﬁnd n > 2 n(n − 1)an, and hence an 6 n−1 . √ Take square roots: 0 < a < √ 2 . n n−1

By the Squeeze Law: limn→∞ an = 0.

The deﬁnition is identical (modulus replaces absolute value). Deﬁnition

The sequence {zn} converges to the complex number z, in symbols, lim zn = z n→∞ means: for every positive there is a natural number N such that for all n > N one has |zn − z| < .

If z ∈ C then n limn→∞ z = 0 if |z| < 1 n limn→∞ z does not exist if |z| = 1 and z 6= 1 n limn→∞ z = ∞ if |z| > 1

limn→∞ zn = ∞ means: for every positive M there is a natural number N such that for all n > N one has |zn| > M.

n Oh yes: limn→∞ 1 = 1.

K. P. Hart Complex power series: an example Introduction Convergence of sequences Convergence of series Important example Sequences of functions Absolute convergence Power series The Logarithm Series

Given a sequence {zn} what does (or should)

z0 + z1 + z2 + z3 + ···

mean? Make a new sequence {sn} of partial sums: X sn = zk k6n

K. P. Hart Complex power series: an example Introduction Convergence of sequences Convergence of series Important example Sequences of functions Absolute convergence Power series The Logarithm Convergence

P If σ = limn sn exists then we say that the series zn converges P and we write σ = n zn.

Thus we give a meaning to z0 + z1 + z2 + z3 + ··· :

the limit (if it exists) of the sequence of partial sums.

This deﬁnition works for real and complex sequences alike.

2 3 n Fix z and consider 1 + z + z + z + ··· (so zn = z ).

The partial sums can be calculated explicitly:

1 − zn+1 s = 1 + z + ··· + zn = (z 6= 1) n 1 − z

for z = 1 we have sn = n + 1.

The limit of the sequence of partial sums is easily found, in most cases: P n 1 |z| < 1: n z = 1−z P n |z| > 1: n z = ∞ also if z = 1

if |z| = 1 and z 6= 1 then the limit does not exist but we do have

2 |s | n 6 |1 − z|

so for each individual z the partial sums are bounded

the bound is also valid if |z| < 1.

P 1 1 n n = ∞ (even though limn n = 0) P (−1)n+1 n n = ln 2 (as we shall see later) P 1 π2 n n2 = 6 (Euler) P 1 n n! = e P 1 n np converges iﬀ p > 1

P Absolute convergence: n |zn| converges.

Absolute convergence implies convergence (but not necessarily conversely).

P (−1)n P 1 n n converges but n n does not

P zn We shall see: n n converges for all z with |z| = 1 and z 6= 1.

comparison if |zn| 6 an for all n P and n an converges P then n zn converges absolutely

K. P. Hart Complex power series: an example Introduction Convergence of sequences Convergence of series Pointwise convergence Sequences of functions Uniform convergence Power series The Logarithm Pointwise convergence

A sequence {fn} of functions converges to a function f (on some domain) if for each individual z in the domain one has

lim fn(z) = f (z) n→∞

n Standard example: fn(z) = z on D = {z : |z| < 1}. We know limn fn(z) = 0 for all z ∈ D, so {fn} converges to the zero function.

fn(z) → f (z) uniformly if for every > 0 there is an N() such that for all n > N() we have

fn(z) − f (z) <

for all z in the domain.

Important fact: if fn → f uniformly and each fn is continuous then so is f .

We have zn → 0 on D but not uniformly: let = 1 , q 2 n 1 1 for every n let zn = 2 then |fn(zn) − f (zn)| = 2 .

Let r < 1 and consider Dr = {z : |z| < r}; n then z → 0 uniformly on Dr . N Given > 0, take N such that r < , then for n > N and all z ∈ Dr we have n n N |z | 6 r 6 r <

P∞ n 1 n=0 z = 1−z for each z ∈ D but not uniformly:

n ∞ X 1 X zn+1 zk − = zk = 1 − z 1 − z k=0 k=n+1

For each individual n this diﬀerence is unbounded.

On a smaller disk Dr we have

n n+1 n+1 X k 1 z r z − = 1 − z 1 − z 6 1 − r k=0

So, on Dr the series does converge uniformly.

P Very useful test: if there is a convergent series n Mn such that

|fn(z)| 6 Mn P for all z in the domain, then n fn converges absolutely and uniformly on the domain.

n n Previous example: |z | 6 r for all z ∈ Dr .

K. P. Hart Complex power series: an example Introduction Convergence of sequences Radius of convergence Convergence of series Boundary behaviour Sequences of functions Summation by parts Power series Back to the boundary The Logarithm Power series

Special form: a ﬁxed number z0 and a sequence {an} of numbers n are given. Put fn(z) = an(z − z0) , we write

∞ X n an(z − z0) n=0 for the resulting series.

Important fact: n P if limn an(w − z0) = 0 for some w then n an(z − z0) converges absolutely whenever |z − z0| < |w − z0|.

Use comparison test: n ﬁrst ﬁx N such that |an(w − z0) | 6 1 for n > N. Then n n n n z − z0 z − z0 an(z − z0) = an(w − z0) 6 w − z0 w − z0 Geometric series with r < 1.

Even better: if r < |w − z0| then the power series converges uniformly on the disc

D(z0, r) = {z : |z − z0| 6 r} Same proof gives n n r an(z − z0) 6 |w − z0| for all z in the disc, apply the M-test.

Theorem P n Given a power series n an(z − z0) there is an R such that P n n an(z − z0) converges if |z − z0| < R P n n an(z − z0) diverges if |z − z0| > R In addition: if r < R then the series converges uniformly on {z : |z − z0| 6 r}.

On the boundary — |z − z0| = R — anything can happen. R = 0, 0 < R < ∞ and R = ∞ are all possible. R is the radius of convergence of the series.

P n n z : R = 1 P 1 n n n z : R = 1 P n n nz : R = 1 P 1 n n n! z : R = ∞ P n n n n z : R = 0 n In each case consider limn anz for various z.

P n P n The series n z and n nz both have radius 1.

What happens when |z| = 1?

The series diverges for all such z n n as neither limn z nor limn nz is ever zero.

P 1 n The series n n z has radius 1.

What happens when |z| = 1?

That depends on z.

P∞ 1 For z = 1 we have divergence: n=1 n = ∞.

P2k 1 1 Remember: n=1 n > 1 + 2 k for all k

P 1 n The series n n z has radius 1.

What happens when |z| = 1 and z 6= 1?

The series converges.

This will require some work.

Remember integration by parts: R fg = Fg − R Fg 0.

The same can be done for sums: let {an} and {bn} be two Pl sequences. We ﬁnd a similar formula for n=k anbn.

The integral is replaced by the sequence of partial sums: Pn An = k=0 ak (and A−1 = 0).

The derivative is replaced by the sequence of diﬀerences: {bn+1 − bn}

The formula becomes

l l−1 X X anbn = Al bl − Ak−1bk − An(bn+1 − bn) n=k n=k The proof consists of some straightforward manipulation.

n 1 1−zn+1 We use this with an = z and bn = n , so An = 1−z

Fix some k and let l > k be arbitrary.

l l−1 X 1 1 1 X 1 1 zn = A − A − A − n l l k−1 k n n + 1 n n=k n=k l−1 ! 2 1 1 X1 1 + + − 6 |1 − z| l k n n + 1 n=k 2 2 = |1 − z| k

This holds for all l, so . . .

Pk 1 n . . . the partial sums n=1 n z form a Cauchy-sequence.

The completeness of the complex plane ensures that

∞ k X 1 X 1 zn = lim zn n k→∞ n n=1 n=1 exists. We denote the sum, for now, by σ(z).

The series converges for all z with |z| 6 1 and z 6= 1.

The inequality l X 1 4 zn n 6 k|1 − z| n=k holds for every z with |z| 6 1 and z 6= 1.

This implies uniform convergence on sets of the form

Er = {z : |z| 6 1, |1 − z| > r}

For z ∈ Er we have

k ∞ X 1 X 1 4 4 σ(z) − zn = zn n n 6 (k + 1)|1 − z| 6 (k + 1)r n=1 n=k+1

4 Now, given > 0 we take n so large that (N+1)r < .

Then σ(z) − Pk 1 zn < whenever k N and z ∈ E . n=1 n > r

K. P. Hart Complex power series: an example Introduction Convergence of sequences Convergence of series Integration Sequences of functions Values on the boundary Power series The Logarithm Integrating the geometric series

We know ∞ X 1 zn = (|z| < 1) 1 − z n=0 We also know k 1 X zk+1 = zn + 1 − z 1 − z n=0 Integrate this along the straight line L from 0 to z:

k X 1 Z w k+1 − Log(1 − z) = zn+1 + dw n + 1 1 − w n=0 L

We can ﬁnd an (easy) upper bound for the absolute value of the integral:

Z k+1 k+1 k+2 w |z| |z| dw 6 |z| × = L 1 − w 1 − |z| 1 − |z| Thus Z w k+1 lim dw = 0 k→∞ L 1 − w and so . . .

. . . we obtain ∞ X 1 zn = − Log(1 − z)(|z| < 1) n n=1 but, by continuity of the sum function σ(z) and the Logarithm this formula holds when |z| = 1 and z 6= 1 as well.

Often z is replaced by −z and an extra minus sign is added to give

∞ X (−1)n+1 zn = Log(1 + z) (|z| 1, z 6= −1) n 6 n=1

K. P. Hart Complex power series: an example Introduction Convergence of sequences Convergence of series Integration Sequences of functions Values on the boundary Power series The Logarithm ln 2

As promised: ∞ X (−1)n = − Log 2 = − ln 2 n n=1 (use z = −1) or, with an extra minus sign:

∞ X (−1)n+1 = ln 2 n n=1 this is also written 1 1 1 1 1 − + − + − · · · = ln 2 2 3 4 5

If z = eiθ, with θ 6= 2kπ, then

∞ X 1 einθ = − Log(1 − eiθ) = − ln |1 − eiθ| − i Arg(1 − eiθ) n n=1

If we split the series and its sum into their respective real and imaginary parts we get two nice formulas.

K. P. Hart Complex power series: an example Introduction Convergence of sequences Convergence of series Integration Sequences of functions Values on the boundary Power series The Logarithm Real part

Note 1 |1 − eiθ|2 = (1 − eiθ)(1 − e−iθ) = 2 − 2 cos θ = 4 sin2 θ 2

iθ 1 So that − ln |1 − e | = − ln 2| sin 2 θ| and we get

∞ X cos nθ 1 = − ln2| sin θ| n 2 n=1

To see what ϕ = − Arg(1 − eiθ) is draw a picture

eiθ ϕ

1 − eiθ

We have 1 − eiθ = (1 − cos θ) − i sin θ, so that if 0 < θ < π we get

sin θ 2 sin 1 θ cos 1 θ cos 1 θ 1 1 tan ϕ = = 2 2 = 2 = tan( π − θ) 1 − cos θ 2 1 1 2 2 2 sin 2 θ sin 2 θ

1 1 1 If 0 < θ < π then ϕ and 2 π − 2 θ lie between 0 and 2 π so that 1 1 ϕ = π − θ 2 2

We ﬁnd ∞ X sin nθ 1 1 = π − θ (0 < θ < π) n 2 2 n=1 The sum must be an odd function so ∞ X sin nθ 1 1 = − π − θ (−π < θ < 0) n 2 2 n=1

K. P. Hart Complex power series: an example