10.1 Taylor expansion of an analytic function
Def. Given a complex-valued function f (z) ; we say that the
”Taylor series of function f at a” is denoted as X1
n=0
f[n] (a)
n! (z a)n : Discussion: If a function f (z) =
X1 n=0
cn (z a)n on some neighborhood of a; we nd that
f0 (z) =
Complex Analysis 132 So we may deduce that ck = f[k] (a) =k!:
Theorem 70 If f is an analytic function on the open disk K = fzj jz aj < Rg ; then f has a Taylor series expansion at a at every point of the disk K, i. e. the power series
X1 n=0
f[n] (a)
n! (z a)n converges to f (z) at every point of K:
pf. To claim that 8 2 K;
Given any point on jz aj < R; take a circle C centered at a with radius r and enclosing : Then 8z 2 C; we have
Complex Analysis 133
Hence the series on the right hand side of equation ( ) converges uniformly and By Cauchy's integral formula, we conclude that
f ( ) =
X1 n=0
f[n] (a)
n! ( a)n ; 8 2 K:
Complex Analysis 134 Theorem 71 (Cauchy's inequality) If f is analytic and
f (z) =
X1 n=0
cn(z a)n
on jz aj < R: Then for a circle C centered at a with radius 0 6= r < R; we nd that
jcnj M
rn; 8n = 0; 1; 2; :::;
where M = max
z2C jf (z)j :
pf. From the previous theorem and Cauchy's integral formula,
cn = 1 2 i
I
C
f (z)
(z a)n+1dz = f[n] (a) n! : Set M = max
z2C jf (z)j ; we can see that jcnj 1
2
M
rn+1 2 r = M rn; 8n = 0; 1; 2; :::
Ex. Expand the function f (z) = fln zg0; 0 < arg z < 2 as a Taylor series at a = 1 + i: Also nd the radius of convergence.
Sol. Note that the function in question is fln zg0 : Cn f0g [ R+ ! E
Complex Analysis 135 , where E = fzj 0 < arg z < 2 g : Then
fln zg0 =
X1 n=0
fln zg[n]0
z=1+i
n! [z (1 + i)]n ; where
fln zg0jz=1+i = ln j1 + ij + i arg j1 + ij = 1
2 ln 2 + 4i;
fln zg00 z=1+i = 1
z z=1+i ; fln zg000 z=1+i = 1
z2 z=1+i ; :
fln zg[n]0
z=1+i = ( 1)n+1(n 1)!
zn
z=1+i
: So we get
fln zg0 = 1
2 ln 2 +
4i + 1
1 + i [z (1 + i)]
+ 1
2 (1 + i)2 [z (1 + i)]2 + ::::
+ ( 1)n+1
n (1 + i)n [z (1 + i)]n + :::::
Complex Analysis 136 But the series on the right hand side has radius of convergence
R = lim
n!1
j1= [n (1 + i)n]j 1= h
(n + 1) (1 + i)n+1i = j1 + ij = p 2:
The domain of f (z) + jz aj < p
2: •
Ex. fln zg0 = ln jzj + i arg z; < arg z < : If we want analyticity at the point z = 1; then we have to choose the branch cut somewhere else, e. g. fzj z 2 R [ f0gg :
Figure.
Then the Taylor expansion of fln zg0 at z = 1 can be expressed as
fln zg0 = (z 1) (z 1)2
2 + (z 1)3
3 +
:::: + ( 1)n 1 (z 1)n
n + ::::;
with jz 1j < 1: Similarly, if we want the analyticity at z = 0; set ez = z + 1; then we may choose the branch cut as fezj ez 2 R [ f0gg ; or simply fz 1g :
Complex Analysis 137 Figure.
The Taylor series expansion for fln ezg0 at ez = 1 can be expressed as
fln ezg0 = ln (z + 1) = z z2
2 + z3 3 + :::: + ( 1)n 1 zn
n + ::::;
with jzj < 1:
Theorem 72 Every bounded entire function is a constant.
pf. 9M(> 0) such that jf (z)j M; and f (z) is analytic, 8z 2 C:
Then for all R > 0;
f (z) =
X1 n=0
f[n] (0)
n! zn; jzj < R;
where
f[n] (0)
n! = cn = 1 2 i
I
jzj=R
f (z)
zn+1 dz:
Complex Analysis 138 According to Cauchy's inequality, we have
jcnj M
Rn; n = 0; 1; 2; ::::
It is obvious that when R ! 1; M=Rn ! 0 if n > 1: Thus jcnj = 0; i. e. cn = 0; 8n 2 N: Therefore we conclude that
f (z) = f (0) ; 8z 2 C;
i. e. f is a constant.
Complex Analysis 139 10.2 Uniqueness theorem
Theorem 73 Given 2 series P1
n=0 an (z z0)n and P1
n=0bn(z z0)n with radii of convergence R1 and R2; re-spectively. If
X1 n=0
an (z z0)n =
X1 n=0
bn (z z0)n on jz z0j < r;
where r = minfR1; R2g ; then an = bn; 8n; and therefore R1 = R2:
pf. Let f (z) denote the sum of P1
n=0an (z z0)n on jz z0j < R1 and g (z) denote the sum of P1
n=0bn (z z0)n on jz z0j < R2. Then obviously,
an = f[n] (z0)
n! ; bn = g[n] (z0) n! :
According to the hypothesis, f (z) = g (z) on jz z0j < r; so then
f[n] (z) = g[n] (z) on jz z0j < r;
and hence
f[n] (z0) = g[n] (z0) on jz z0j < r; 8n;
i. e. an = bn; for all numbers of n:
Complex Analysis 140 Theorem 74 (Uniqueness theorem for power series)
X1
pf. Choose a sequence f kg of distinct points of E such that
klim!1 k = z0; but k 6= z0:
Note that this hypothesis is designed by the de nition of a limit point. Then we may see that
klim!1 holds for all m 2 N; so we have proved the theorem by induction.
Theorem 75 (Uniqueness theorem for analytic functions)
f (z) and g (z) are both analytic functions on the same do-main G; and f (z) = g (z) on E G; where E contains a limit
Complex Analysis 141 point z0 on G: Then f (z) g (z) on G:
pf. 8& 2 G; (& 6= z0; ) there exists certain continuous curve C contained on G and C joins z0 and &:
De ne be the distance between C and the boundary of G if G is bounded, otherwise let be any positive number. Select a set of points fzkgnk=0 on curve C such that jzk zk+1j < ; k = 0; 1; 2; :::n 1; zn = &; and all of the disks Kk = fzj jz zkj < g ; lie entirely on G:
Figure.
Now, write f (z) and g (z) as series on K0; i. e.
f (z) =
X1 n=0
an(z z0)n; and g (z) =
X1 n=0
bn(z z0)n:
Since f = g on E \ K0; while z0 is also a limit point of E \ K0: By the previous theorem, an = bn on K0; 8n; i. e.
f (z) = g (z) on K0:
Complex Analysis 142 Next, write f (z) and g (z) as series on K1; i. e.
f (z) =
X1 n=0
a0n(z z1)n; and g (z) =
X1 n=0
b0n(z z1)n: Then again, since f = g on K0 \ K1; while z1 is a limit point of K0 \ K1: Also by the previous theorem,
f (z) = g (z) on K1:
Keep repeating the same process, we nd that f (&) = g (&) on Kn+1: Because & and the curve C are arbitrary on G; it is natural to conclude that f (z) = g (z) on G:
Def. Any root of f (z) = 0 is called a zero of f:
Def. f (z) ( = 0) is analytic on a domain G and has a zero z0 on G: If
f (z) =
X1 n=m
cn (z z0)n;
where m 1 and cm 6= 0; then we say that z0 is a ”zero of order m:” When m = 1; z0 is also called a ”simple zero of f (z);” and z0 is called a ”multiple zero of f (z)” if m > 1:
Theorem 76 Every zero z0 of an analytic function f (z) on a do-main G is isolated.
pf. Suppose that every neighborhood K of z0 contains some other zero z 6= z0: Let E is the collection of all of those zeros,
Complex Analysis 143 then clearly z0 is a limit point of E: According to the uniqueness theorem of analytic function, we have
f (z) 0 on G; (! ) :
Theorem 77 Any analytic function of constant modulus is itself constant.
pf. jf (z)j = M : a constant. Since f is analytic, consider the following two cases:
i). M = 0 :
From the previous theorem, we have f (z) 0: ] ii). M 6= 0 :
Consider the function f (z) = u (z) + iv (z) ; z = x + iy;
note that
0 6= M2 = u2 + v2: Since f is analytic, we may see
2uux + 2vvx = 0 and 2uuy + 2vvy = 0;
where u and v are to be determined. Since u and v are not identically zero, the determinantof the above equations
ux; vx
uy; vy = 0;
i. e. uxvy uyvx = 0: By Cauchy-Riemann equations, we realize
Complex Analysis 144 that
u2x + vx2 = 0 and u2y + vy2 = 0:
Therefore
ux = vx = uy = vy = 0;
i. e. u = c1 and v = c2 : constants. Thus f (z) is a constant.
Complex Analysis 145 10.3 The maximum modulus principle and its applications Def. h (z) be a continuous function on a domain D: Let z0 2 D; value of its values over any small circles centered at z0: i. e.
8z0 2 D; 9" > 0 such that
Theorem 78 If u (z) is a harmonic function on a domain G; and if the disk Kz0 = fzj jz z0j < g is contained on G; then pf. We begin with the identity
0 =
which follows immediately from Green's theorem and the
Complex Analysis 146 harmonicity of u (z) : We parametrize the circle by
x ( ) = x0 + r cos and y ( ) = y0 + r sin ; Since u is smooth, the above equation becomes
0 = r
Remark 18 i). A harmonic function must have the mean value property.
Complex Analysis 147 ii). An analytic function is composed by two harmonic func-tions and thus also have the mean value property.
Ex. Show the mean value property of analytic function directly without invoking harmoniticity and Green's theorem.
(Homework)
Theorem 79 (Maximum modulus principle)
f (z) is analytic and non-constant on a domain G; then jf (z)j cannot have a maximum on G:
pf. Suppose that 9z0 2 G such that jf (z0)j jf (z)j, 8z 2 G: Let M = jf (z0)j > 0; otherwise, f (z) 0 on G: Consider a circle R centered at z0 with radius R such that
R G; then by the mean value property, we have f (z0) = 1
2
Z 2 0
f z0 + Rei d ; and
M = jf (z0)j 1 2
Z 2 0
f z0 + Rei d : Since M f z0 + Rei ; 8 ; also note that
M = 1 2
Z 2 0
M d :
Complex Analysis 148 Shift the right hand side to the left,
0 1
2
Z 2 0
M f z0 + Rei d ; i. e. the integral is 0: Hence we must have the result that
M = f z0 + Rei ; 8 ;
or jf (z)j = M on the circle R: Again, since R is arbitrary, which means for any open disk Kz0 centered at z0; we may always nd a circle R inside of Kz0 and get the same result.
i. e. jf (z)j = M on Kz0: By uniqueness theorem for analytic functions, we conclude that f (z) must be a constant on G; which is a contradiction.
Corollary 80 (Minimum modulus principle)
An analytic, non-constant and non-vanishing functionf (z) on a domain G does not have a minimum modulus.
pf. De ne (z) = 1=f (z) : Then obviously, (z) is also analytic, non-constant and non-vanishing on G: By maximum modulus principle, j (z)j doesn't have a maximum on G;
thererfore we nd that jf (z)j doesn't have a minimum on G:
Corollary 81 f (z) is an analytic function de ned on a bounded domainG and continuous on G: Then9 2 @G such that jf ( )j = maxz2G jf (z)j :
Complex Analysis 149 pf. We may discuss the problem in the following two
conditions:
i). f (z) is constant on G:
Then can be chosen to be any point on @G:
ii). f (z) is not a constant on G:
Then f (z) must be non-constant on G; otherwise, the whole f would become constant. Again, use maximum modulus principle, f cannot have maximum on G; so the maximum value of jf (z)j must happen somewhere on @G:
Corollary 82 f (z) is an analytic and non-vanishing function de-ned on a bounded domain G and continuous on G: Then 9 2
@G such that jf ( )j = minz2G jf (z)j :
Corollary 83 f (z) is an analytic and non-vanishing function de-ned on a bounded domain G and continuous on G: If f (z) is a constant on @G; then f (z) is constant everywhere on G:
pf. By maximum and minimum modulus theorems, we nd that
M = min
z2G jf (z)j = max
z2G jf (z)j :
i. e. jf (z)j = M for all z 2 G: Obviously, jf (z)j = M for all z 2 G; so we nd that f (z) is in fact a constant on G: Again, since f (z) is continuous on G; we conclude that f (z) must also
Complex Analysis 150 be the same constant on @G; and hence f (z) is a constant on the whole G:
Next, we shall use the analogy about analytic functions to harmonic functions.
Theorem 84 u (z) is harmonic and non-constant on a domain G:
Then u (z) has neither a maximum nor a minimum value on G:
pf. 8z0 2 G; take a neighborhood K of z0 such that K G:
Sice K is simply connected, we may nd a conjugate harmonic v of u such that an anlytic function
f (z) = u (z) + iv (z) on K:
De ne g (z) = exp f (z) ; which is analytic on K and non-constant. Then jg (z)j = exp u (z) 6= 0: Use maximum (& minimum) modulus principle, there is no maximum (or minimum) value(s) on K; and hence there is no maximum (or minimum) value of u (z) on K:
Because z0 is arbitrary on G; we conclude that u(z) does not have a maximum (or minimum) value on G:
Theorem 85 u (z) is harmonic on a bounded domain G and con-tinuous onG: Then 9 m, M 2 @Gsuch that u ( m) = minz2Gu (z) and u ( M) = maxz2G u (z) :
pf. We may consider the problem in the following two cases:
Complex Analysis 151 i). u (z) is a constant on G :
Then m, M can be chosen as any two points among @G:
ii). u (z) is non-constant on G :
u (z) has no maximum (nor minimum) value on G; yet the maximum and the minimum values must occur somewhere on G because of the continuity of u: Therefore they can only occur on
@G:
Theorem 86 u (z) is harmonic on a bounded domain G and con-tinuous on G: Then if u (z) is constant on @G; u (z) is constant everywhere on G:
Theorem 87 u1 (z) ; u2(z) are harmonic on a bounded domain G and continuous on G: If u1 (z) = u2(z) on @G; then u1(z) = u2 (z) on G:
Complex Analysis 152