Complex Integrals Integral : Definite: It is regarded as the limit of a sum. Indefinite: It is the process inverse to differentiation. Continuous Arc: If a point z on an arc is such that z= + …. (i) and z=x+iy. We may write, x= = ……. (ii). If are real valued and continuous function of the real variable t defined on the interval a≤t≤b then the arc is called continuous arc. Multiple Points: If the relation x= = are satisfied by two or more values of t in the given interval then the point z=(x,y) is called a multiple point. Jordan Arc/Simple Arc: A continuous arc having no multiple points is called Jordan arc. i.e. we say L is simple or Jordan arc if z(t1)=z(t2) only when t1=t2. Regular arc of a Jordan Curve: Considering an arc of a Jordan curve defined by the equation z= + where a≤t≤b. If z can be expressed as a single valued and , as well as % , % are continuous in the interval a≤t≤b, the arc then is called regular arc of a Jordan curve. Contour: Contour A contour or piece wise smooth arc, is an arc consisting of a finite number of regular arc joined end to end. Hence if z=zt represent a contour; zt is continuous where as derivative z’t is piecewise continuous. continuous x + ix when 0 ≤ x ≤ 16 For example: z=2 is contour. x + i when 1 ≤ x ≤ 2 When only the initial and final values of z(t) are same a contour c is called a simple closed contour. The unit circle z=e78 (0 ≤ 9 ≤ 2:) is closed contour. Rectifiable Arc: Let L be an arc defined by z=z(t)=x(t)+iy(t) where a≤t≤b. Let us consider a partition P={t0,t1,…tn} of [a,b]. Corresponding to the partition, let L be subdivided into n-subarcs where zi=zti. We join each of the points z0, z1, …zn to the next point by the line segment then the length of this polygon is ∑LHMK|GH − GHJK|. We say that the arc L is rectifiable if the least upper bounds (or supremum) of the sums |GK − GN| + |GO − GK| + ⋯ + |GL − GLJK |…..(i) taken over all partition P is finite. i.e. if Sup∑LHMK|GH − GHJK| = Q < ∞. Also we say that L is non-rectifiable if the sum (i) becomes arbitrarily large for suitably choosen partition. In this case we say that L has no length (or infinite length). Contour Integrals (Line Integrals) Suppose that the equation z=z(t), a≤t≤b represents a contour c, extending from a point z1=za to a point z2=zb. Let fz be piece wise continuous on the interval a≤t≤b we define the line integral or contour integrals of f along c as follows, Y
TV UG G = TZ UWGXG %
[\]
[dz=
[]
. = G %. ] 1
In the form of limit sum we can define the line integral as:
∫
c
n
f ( z ) dz = lim ∑ f (ξi )( zi − zi −1 ) where zi-1≤ξi≤zi n →∞
i =1
Where the existence of the integral is ensured by the piecewise continuity of z’(t). Further it is easy to see that T_ UG G = − T_ UG G
i.
If C=c1 +c2 then T_ UG G = TV UG G + TV UG G
ii.
`
T_ GNUG G = GN T_ UG G
iii.
a
T_ [UG + bG] G = T_ UG G + T_ bG G
iv.
∫
v.
b
b
f ( z )dz ≤ ∫ f (z(t)) z ' (t ) dt. If f ( z ) ≤ m and ∫ z ' (t ) dt = L
c
a
a
b
∫ f ( z)dz ≤ ∫ c
b
f ( z (t )) z ' (t ) dt ≤ m ∫ z ' (t ) dt = mL
a
a
Example: 1. Using the definition of an integral as the limit of a sum evaluate the following integrals. i.
∫ dz
∫
ii.
l
dz
l
iii.
∫ zdz l
where ℓ is any rectifiable arc joining the points z=a and
z=b. n
Solution: i. By the definition of complex integration; we have
i.e.
∫ dz = lim( z − z + z − z + ..... + z ⇒ ∫ dz = lim( z − z ) = b − a. n →∞
l
1
0
n
n →∞
l
2
1
ℓ
l
n →∞
i =1
− zn−1 + ...)
0
In particular, if ℓ is closed then zn=z0 i.e.
∫
n
∫ dz = lim ∑ ( zi − zi−1 )
∫ dz = 0 ℓ
n
dz = lim ∑ zi − zi −1 = lim z1 − z0 + z2 − z1 + ...... + zn − zn −1 n →∞
i =1
n →∞
ii. Solution: = lim [ chordz1 z0 + chordz2 z1 + .... + chordzn zn −1 ] n →∞
= arcz1 z0 + arcz2 z1 + ..... + arczn zn −1
= arc length ℓ.
∫
iii. Here,
ℓ
n
zdz = lim ∑ ξ i ( zi − zi −1 )......(1) where ξi is any point on the sub-arc joining zi-1 n →∞
i =1
to zi. Since ξi is arbitrary, we choose ξi =zi and ξi =zi-1 successively in (1) we get
∫
ℓ
n
zdz = lim ∑ zi ( zi − zi −1 )......(2) and n →∞
i =1
Adding (2) and (3) we have,
2
∫
ℓ
n
zdz = lim ∑ zi −1 ( zi − zi −1 )........(3) n →∞
i =1
n
2 ∫ zdz = lim ∑ { zi ( zi − zi −1 ) + zi −1 ( zi − zi −1 )} n →∞
ℓ
i =1
n
⇒ 2 ∫ zdz = lim ∑ ( zi2 − zi2−1 ) = lim( z12 − z02 + z22 − z12 + ...... + zn2 − zn2−1 + ...) n →∞
ℓ
n →∞
i =1
⇒ 2 ∫ zdz = lim( zn2 − z02 ) = (b 2 − a 2 ) n →∞
ℓ
If ℓ is closed then
∫ zdz = 0 ℓ
2. Find the value of the integral, I= zdz where C is given by z= 2eiθ (−
∫ c
π 2
≤θ ≤
π 2
) of the
circle z = 2 from z=-2i to 2i. For, z= 2eiθ , When,
π θ =− , 2
π
π
z = 2 × cos(− ) + i sin(− ) = 2 × (0 − i ) = −2i 2 2
Whenθ =
π
2
,
π
c
π
z = 2 × cos( ) + i sin( ) = 2 × (0 − i ) = 2i 2 2 Now,
I = ∫ f ( z )dz = c
=
π /2
∫ π
π /2
∫ π
f ( z (θ )) z ' (θ )dθ
− /2
( z (θ ))2ieiθ dθ =
− /2
π /2
∫ π
2e − iθ 2ieiθ dθ = 4i
− /2
π /2
∫ π
dθ =4i[θ ]π−π/ 2/2
− /2
= 4π i Further we can find the integral
1
∫ z dz
for z=-2i to 2i
We have, z = 2, z = 4 i.e. z z = 4 , z = 2
4 z
Therefore, we have,
∫
zd z = 4π i
i e .∫ c
∴
∫ c
4 dz = 4π i z 1 dz = π i z
Example:2 Find
∫ f ( z )dz where f(z) =y-x-i3x
2
and c=c1-c2 as required in the figure.
c
3
A
B C2
C1 O
∫ f ( z)dz = ∫
Here,
c1
f ( z )dz +
OA
∫
f ( z )dz
AB
The leg OA may be represented parametrically as z=0+iy (0≤y≤1) and since x=0 at point on that leg, the values of f vary with the parameter y according to the equation f(z)=y (0≤y≤1).
∫
Now,
OA
So,
1
1
0
0
f ( z )dz = ∫ f ( z ( y )) z ' ( y )dy = ∫ yidy where f(z)=y-x-i3x2, f(0+iy)=y-0=y, z’(y)=i. 1
∫
f ( z ) dz = ∫ yidy = i[
OA
0
y2 1 1 i ]0 = i × = . 2 2 2
Next, on the leg AB, z=x+i (0≤x≤1) and So,
∫
1
f ( z )dz = ∫ f ( z ( x)).z ' ( x) dx.
AB
f(z(x))=y-x-i3x2, but z(x)=x+i. Then f(z(x))=1-x-i3x2.
So, So.
0
∫
1
f ( z )dz = ∫ [1 − x − i3 x 2 ]1.dx where z’(x)=1.
AB
0
1
1
1
0
0
0
= ∫ dx − ∫ xdx − i3∫ x 2 dx = [ x]10 − [
Therefore,
∫ f ( z )dz = ∫ c1
f ( z )dz +
OA
x2 1 x3 1 1 ]0 − 3i[ ]10 = 1 − − i1 = − i . 2 3 2 2
∫
f ( z )dz =
AB
i 1 1− i + −i = . 2 2 2
If C2 denotes the segment OB of the line y=x with parametric representation z=x+ix (0≤x≤1). So,
∫
1
1
0
0
f (z) dz = ∫ f ( z (x)) z ' ( x) dx = ∫ ( x − x − i3 x 2 )(1 + i ) dx
OB
1
∫
= −i3 x 2 dx(1 + i ) = (1 + i ) − i3[ 0
Hence,
∫
c = c1 − c2
x3 1 ]0 = −i[1 + i ].1 = 1 − i 3
f ( z )dz = ∫ f ( z )dz − ∫ f ( z )dz. c1
In the closed contour OABO,
c2
∫ f ( z )dz = c
1− i i −1 − (1 − i) = 2 2 4
Anti-derivatives: Let f be continuous function on a domain D, then an anti-derivative F of f is a function s.t. F’(z)=f(z) for all z∈D. Theorem: Suppose that a function f is continuous on a domain D. If any one of the following statement is true, then so are the others: a. f has an anti-derivative in D. b. The integral of f(z) along contours lying entirely on D and extending from any fixed point z1 to any point z2 all have the same value. c. The integrals of f(z) around closed contours lying entirely in D all have zero. Proof: Suppose that statement (a) is true. If a contour c from z1 to z2 lying on D is smooth are with parametric representation z=z(t) (a≤t≤b) . We know that,
∫
Now,
dF ( z (t )) = F ' ( z (t )) z '(t ) = f ( z ( t )).z ' ( t ) ,(a ≤ t ≤ b) dt b
b
a
a
f ( z ) dz = ∫ f ( z (t )) z '(t) dt = ∫
c
d F ( z (t )) dt = [ F ( z (t ))]ba dt
= F ( z (b)) − F ( z ( a )) = F ( z2 ) − F ( z1 ). Furthermore if C consists of a finite number of smooth arcs Ck (k=1,2,3,…,n-1) each Ck extending from a point zk to the point zk+1. Then,
∫
n −1
n −1
k =1 ck
k =1
f ( z )dz = ∑ ∫ f ( z )dz =∑ [F(z k +1 ) − F ( zk )]
c
= F ( z2 ) − F ( z1 ) + F ( z3 ) − F ( z2 ) + ... + F ( zn ) − F ( zn −1 ) = F ( zn ) − F ( z1 ) . Hence ⇒ f . Next we show that (b) implies (c). Since (b) is true, we have
∫ f ( z )dz = ∫ f ( z )dz where c
1
c1
and c2 are closed contour acting from z1 to z2. So,
∫ f ( z)dz − ∫ f ( z )dz = 0 c1
c2
∫ f (z) dz+ ∫
− c2
c1
Or,
f ( z )dz = 0
∫
f ( z )dz = 0
c = c1 − c2
i.e.
∫
c = c1 − c2
f ( z )dz = 0 . It follows that (b) implies (c).
Finally we show that ©implies (a). 5
c2
z
We can define a function F=
∫ f (s)ds. z0
Clearly, the assertion [c⇒a] will be proved if we can show F’(z)=f(z), ∀z ∈ D . We do this by letting z+△z be any point distinct from z, lying in some nbd of z that is sufficiently small to be contained in D. Then z +∆z
F(z+△z)=
∫
f ( s )ds and F ( z + ∆z ) − F ( z ) =
z +∆z
z0
F ( z + ∆z ) − F ( z ) 1 ⇒ = ∆z ∆z z +∆z
But
∫
ds = ∆z . So, f(z)=
z
Therefore, Thus, l
h\i∆\Jh\ ∆\
h\i∆\ − ∆\
f ( s )ds − ∫ f ( s )ds =
z +∆z
z0
z0
z
∫
f ( s )ds
z +∆z
∫
f ( s )ds.
z0
f (z) f (z) .∆z = . ∆z ∆z K
\i∆\
− UG = ∆\ T\ K
∫
z
\i∆\
UGl = l T\ ∆\
z +∆z
∫
ds =
z
1 ∆z
z +∆z
∫
\i∆\
K
Uk k − ∆\ T\ K
f ( z )ds.
z
K
\i∆\
UG k = ∆\ T\
WUk − UGX k.
∆\
Uk − UG kl ≤ |∆\| T\ |Uk − UG| k.
Since, f is continuous in D, we have for every ∈> 0 ∃ o > 0 k. . |Uk − UG| < p qℎs stsu |k − G| < o. Here we take |∆G| < o. vws uw , |k − G| < o. l
h\J∆\Jh\ ∆\
x
− UGl < |∆\| |∆G| = y. Which implies that, F’(z)=f(z) ∀ G {.
Thus F is an antiderivative of f. So © implies (a).
Goursat’s Lemma: Let f(z) be analytic throughout the closed region R consisting of the points interior to a positively oriented simple closed contour C together with the point on C itself. For any positive number ϵ, the region R can be covered with a finite number of square and partial squares, indexed by j=1,2,3,..n such that in each one there is a fixed zj for which the inequality
f ( z) − f ( z j ) z − zj
− f '( z j ) < ε (z≠zj)…..(1) is satisfied by all other points in that square or partial
square. Proof: If possible suppose the lemma is false. Then there exist ε>0 s.t. we subdivide the region C, there will be at least one mesh (either square or partial square) for which the statement does not hold. We shall this leads to contradiction.
6
Let the closed regions are consisting of points within and on c be covered by a network of finite no. of meshes by drawing lines parallel to the co-ordinate axis. Then by our assumptions, there is at least one mesh for which (1) is false. Let σ0 denote the mesh of square. We divide σ0 into four equal squares. Then there is at least one of the four smaller square say σ1 containing the points of R for which (1) does not hold. We further divide σ1 into four square and this process of division continued further. If this process ends after a finite number of steps, we have arrieved at a contradiction and our lemma is true. If however the process is continued indefinitely, we get a nested square of squares, σ0, σ1, σ2….. σn each contained in the preceding one for which the lemma is false. The square, therefore, determines a point z0 common to all these squares such that z0 is limit point of the set of point in the closed region R. Since R is closed and z0∈R. Now by hypothesis, f(z) is analytic at z0. Consequently f’(z0) exists. So by definition of derivative, choose ϵ>0 ∃δ>0 s.t.
f ( z ) − f ( z0 ) − f '( z0 ) < ε holds for all z for which z − z0 < δ . z − z0 We can choose a positive N so large the diagonal of the square σn is less than σ. So that for n≥N all
the squares σn are contained in the circular nbd z − z0 < δ of z0. Also z0∈ σN for n. Thus we arrived at contradiction by taking zj to be z0. The inequality is satisfied. Cauchy Goursat Theorem If f(z) is analytic at all points within and on a simple closed contour C. Then
∫ f ( z )dz = 0. . c
Proof: Let ε>0 be given, then by the lemma the region C can be divided into finite number of square of partial squares where zi is the fixed point lying in subregion and appearing in the statement of lemma,
f ( z ) − f ( zi ) − f '( zi ) forz ≠ zi δ i ( z ) = z − zi ......(i) 0 forz = zi Within each δi where δ i ( z ) < ε , ∀ε > 0 ….ii . Also the function δi(z) is continuous throughout the subregion. Since f(z) is continuous there and
lim δ i ( z ) = f '( zi ) − f '( zi ) = 0 . z → zi
Next let Si (i=1,2,3…n) denote the positively oriented boundaries of the above square or partial squares covering the region within C. Then from the definition (i) the value of f at a point z on any particular δ i can be written as f(z)=f(zi) - zif’(zi) + f’(zi)z + (z-zi)δiz and it follows that
7
∫
f ( z ) d z = [ f ( z i ) − z i f '( z i )] ∫ d z + f '( z i ) ∫ zd z +
si
si
. But
i
i
( z ) d z .....( iii )
si
∫ dz = 0 and ∫s zdz = 0 (By Cauchy theorem) si
So,
si
∫ ( z − z )δ
∫
i
f ( z)dz =
si
∫ (z − z
i
) δ i ( z ) d z ......( iv )
si
It is clear the sum of integrals around all Si’s is the integral around the closed contour C. Since the line integrals along the common boundary of every pairs of adjacent meshes cancel each other. Only n
the integral along the arc from the part c remains, then,
∫ f ( z )dz = ∑ ∫ f ( z )dz......(v) i =1 si
C
Thus from (iv) and (v),
∫
f ( z ) dz =
∑ ∫ ( z − z )δ
∫
i
i =1 s i
c
So that we have,
n
f ( z )dz =
n
∑
i =1
c
< ε
n
∑∫ i =1 si
i
( z ) dz
( z − z i )δ i ( z ) d z ≤
.
n
∑∫ i =1 si
z − zi δ i ( z ) d z
z − z i d z .......( v i )
The boundary Si of a mesh either a complete square or a partial square. Since z lies on Si and zi lies either on or inside Si, let ai denotes the length of the sides of a square. The distance between z and zi can not exceed the length ai√2 of the diagonal of that square.
z − zi ≤ ai 2.
So that
Hence,
∫ si
z − zi d z ≤ ai
2 . ∫ d z .....( v ii ) . But si
∫ dz si
represent the length of Si. Here
this length is 4ai if Si is complete square. And it can not exceed (4ai+ℓi) if Si is a partial square, where ℓi is the length of the subarc of C. Hence, if Si is a square, then the inequality (vii) gives,
∫ z−z
i
dz ≤ ai 24a i = 4 2 a i 2 ......(viii)
i
dz ≤ ai 2(4ai + li ) ≤ 4 2ai 2 + 2ai li .....(ix)
and if Si is a partial square then,
si
∫ z−z si
8
Let ‘a’ is the length of a side of square enclosing the entire curve C. It is clear that the sum of area ai2 of all these square cannot exceed a2. Hence if ℓ is the length of curve C then from (vi), (viii) and (ix),
∫
n
f ( z )dz < ε ∑ (4 2ai 2 + 2ai li ) < ε (4 2a 2 + 2al ) = ε k where k is constant. i =1
c
Thus,
∫ f ( z )dz = 0. c
Simply and multiply connected domain: A region R is said to be simply connected region if every closed contour within it encloses only the points of R. In short, region without hole.
. . . ..
If all the points of the region ‘R’ are bounded by two or more closed curves thus the region R is said to be multiply connected region. In short, region with hole.
…
Cross Cuts: The lines in a multiply connected region without intersecting any one of the curves, which makes a multiply connected region a simply connected are called cross cuts. Some consequences of Cauchy Goursat’s Theorem: b
1. If f(z) is analytic in a simply connected region R, prove that
∫ f ( z )dz is independent of the a
path in R joining any tow points a and b in R. Proof: Let c1 and c2 be any two curves in R joining a and b consisting c1 and c2.
∫ f ( z )dz = 0 where c denotes the closed
Then by Cauchy Goursat’s theorem,
c
curves joining a and b. Then by Cauchy Goursats theorem,
∫ f ( z )dz = 0 ⇒ ∫ f ( z)dz + ∫ c1
c
⇒
∫ f ( z)dz − ∫ f ( z )dz = 0 c1
⇒
f ( z )dz = 0
− c2
c2
∫ f ( z )dz = ∫ f ( z )dz c1
c2
2. Let f(z) be analytic in a Region R bounded by two simple closed curves c1 and c2 proved that
∫ f ( z )dz = ∫ f ( z )dz where c c1
1
and c2 are both traversed in the positive sense relative to
c2
their interiors (or, anticlockwise) 9
..
Proof: Let us define the cross cuts joining a point D of a curve c1 to a point E of the curve c2. Then the region formed by DABCDEFGED is simply connected region. Hence by Cauchy Goursat theorem, we have,
∫
DABCDEFGED
∫
Or,
DABCD
∫
But,
DE
∫
DABCD
Or,
∫
c1
f ( z ) dz + ∫
DE
f ( z )dz = ∫
− ED
f ( z )dz + ∫
EFGE
f ( z )dz + ∫
− c2
f ( z )dz = 0
f ( z ) dz + ∫
EFGE
f ( z ) dz + ∫
ED
f ( z ) dz = 0...(i )
f ( z )dz . Hence (i) reduced to, f ( z ) dz = 0
f ( z )dz = 0
∴ ∫ f ( z )dz = ∫ f ( z )dz. c1
c2
Cauchy Integral Formula: If f(z) is analytic within and on a closed contour c and if z0 is any point within c then f(z0)=
1 f ( z )dz ∫ c 2π i z − z0
Proof: Let α be a circle, z − z0 = r with centre z0 where the radius r is small enough so that α is interior to c. Then the function
f (z) is analytic at all points within and on c except at the z − z0
point z0. Hence by Cauchy Goursat theorem for multiply connected region,
∫
c
f ( z ) − f ( z 0 ) + f ( z0 ) f ( z ) dz f ( z ) dz =∫ =∫ dz α α z − z0 z − z0 z − z0
⇒∫
c
f ( z ) − f ( z0 ) f ( z0 ) f ( z ) dz dz + ∫ dz.......(1) =∫ α α z − z0 z − z0 z − z0
Where both integral are taken
counter clockwise. Let us put z-z0= re iθ . So that dz=ireiθdθ. Then,
∫α
iθ 2π rie 2π f ( z0 ) dz = f ( z0 ) ∫ dθ = f ( z0 ) ∫ idθ = 2π if ( z0 ) . iθ 0 0 z − z0 re
10
We now show that the first integral
∫α
point z=z0. There corresponds to each
f ( z ) − f ( z0 ) is zero. Since fz is continuous at the z − z0
ε > 0 ∃ aδ > 0 s .t f ( z ) − f ( z 0 ) < ε
where z − z0 < δ .......(2) . Let us choose any positive number ‘r’ less than δ which is small enough so that the positively oriented circle z − z0 = r is interior to c. Thus f ( z ) − f ( z0 ) < ε where z − z0 = r . Thus
∫α
f ( z ) − f ( z0 ) ε < z − z0 r
ε
∫α dz = r 2π r = ε 2π
Consequently the absolute value of the integral can be made arbitrarily small. The value of the integral must therefore be zero. So the equation 1 reduces to,
∫
c
f ( z ) dz = f ( z0 )2π i z − z0
∴ f ( z0 ) =
1 f ( z ) dz . ∫ c 2π i z − z0
Extension of the Cauchy integral formula for multiply connected region: Let fz be anylatic within and on a multiply connected region enclosed by two closed curves c1 and c2 and if z0 be any point within it then fz0=
1 f ( z) 1 f ( z) dz − dz ∫ ∫ c c 1 2 2π i z − z0 2π i z − z0
Proof: Let us define the cross cuts joining a point D of a curve c1 to a point E of the curve c2. Then the region formed by DABCDEFGED is simply connected region and which is closed contour. Hence by Cauchy integral for any closed contour, we have,
f ( z0 ) =
1 f ( z ) dz ∫ DABCDEFGED 2π i z − z0
Or,
f ( z0 ) =
But,
1 f ( z) 1 f ( z) 1 f ( z) 1 f ( z) dz + dz + dz + dz......(i ) ∫ ∫ ∫ ∫ DABCD DE EFGE ED 2π i z − z0 2π i z − z0 2π i z − z0 2π i z − z0
1 f ( z) 1 f ( z) dz = dz . Hence (i) reduced to, ∫ ∫ DE − ED 2π i z − z0 2π i z − z0
f ( z0 ) =
1 f (z) 1 f ( z) dz + dz ∫ ∫ DABCD EFGE 2π i z − z0 2π i z − z0 11
f ( z0 ) =
1 f ( z) 1 f ( z) dz + dz ∫ ∫ 2π i c1 z − z0 2π i − c2 z − z0
f ( z0 ) =
1 f (z) 1 f ( z) dz − dz ∫ ∫ c c 2π i 1 z − z0 2π i 2 z − z0
We can generalized this theorem for multiply connected region R bounded by simple closed curves c1,c2,……,cn for the point z0 as,
f ( z0 ) =
1 f ( z )dz 1 f ( z ) dz 1 f ( z ) dz − − .... − ∫ ∫ ∫ c c c 2π i 1 z − z0 2π i 2 z − z0 2π i n z − z0
Derivative Derivative of an analytic function Let fz be analytic within and on the boundary c of a simply connected region R and let z0 be any point with in c. Then f '( z0 ) =
1 f ( z ) dz where c is any closed contour in ∫ c 2π i ( z − z0 ) 2
region R. Proof: Let z0+h be any point within R near the point z0 so that h is at our choice. Now by Cauchy integral formula we have, fz0+h=
1 f ( z ) dz 1 f ( z ) dz and f ( z0 ) = ∫ ∫ c c 2π i z − z0 − h z − z0 2π i
Therefore,
f ( z 0 + h ) − f ( z0 ) =
1 1 1 1 hf ( z )dz f ( z) − dz = ∫ ∫ c c 2π i 2π i ( z − z0 − h)( z − z0 ) z − z0 − h z − z0
And also,
f ( z 0 + h ) − f ( z0 ) 1 f ( z ) dz = ∫ c h 2π i ( z − z0 − h)( z − z0 ) =
( z − z0 − h ) + h 1 1 f ( z ) dz hf ( z ) dz f ( z ) dz = +∫ 2 2 ∫ ∫ c c c 2π i ( z − z0 − h)( z − z0 ) 2π i ( z − z0 ) ( z − z0 − h)(z − z 0 ) 2
i.e.
f ( z0 + h ) − f ( z 0 ) 1 f ( z ) dz 1 hf ( z )dz − = ....(1) 2 ∫ ∫ c c h 2π i ( z − z0 ) 2π i ( z − z0 − h)( z − z0 ) 2
The theorem will be proved if the integral of right side of 1 tends to zero as h tends to zero. For this purpose we describe a circle β with centre z0 and radius ρ s.t. β lies entirely within c. Then by consequences of Cauchy Gaursat theorem, we have,
1 hf ( z ) dz 1 hf ( z ) dz = ....(2) 2 ∫ ∫ c β 2π i ( z − z0 ) ( z − z0 − h) 2π i ( z − z0 ) 2 ( z − z0 − h)
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We now choose h such that the point z0+h lies within β and that h <
1 ρ . Now the 2
equation of β is z − z0 = ρ . Then we have, z − z0 − h ≥ z − z0 − h > ρ −
ρ 2
=
ρ 2
.
Since fz is analytic in the region R. By upper boundness we can find m>0 s.t. f ( z ) ≤ m. Now from 1 and 2 we get
f ( z0 + h ) − f ( z0 ) 1 f ( z )dz 1 hf ( z )dz − = 2 ∫ ∫ h 2π i c ( z − z0 ) 2π i β ( z − z0 − h)( z − z0 ) 2 ≤
1 h m ∫β dz 1 h m.2πρ 2 h m = = ....(3) ρ2 2π ρ .ρ 2 2π ρ ρ 2 2 2
Hence when h→0, the right hand side of 3 also tends to zero and we get,
lim h →0
f ( z0 + h ) − f ( z0 ) 1 f ( z ) dz = ∫ c h 2π i ( z − z0 ) 2
Thus fz is differentiable at z0 and f '( z0 ) =
1 f ( z ) dz . 2π i ∫c ( z − z0 ) 2
Analytic character of successive derivative derivative of an analytic function If fz is analytic within and on a boundary c of a simply connected region R and z0 is any point with in c, then fz possess derivative of all orders and these derivative are all analytic at z0 their values being given by f n ( z0 ) =
n! f ( z ) dz ∫ c 2π i ( z − z0 ) n +1
Proof: From previous theorem we have,
f '( z0 ) =
1 f ( z ) dz ∫ c 2π i ( z − z0 ) 2
This shows that the result of the theorem is true for n=1. Assume that the result is true for n=m. So that f m (z 0 ) =
m! f ( z) ∫ c 2π i ( z − z0 ) m +1
Let z0+h be a point in the neighbourhood of z0. Then
f m ( z0 + h) − f m ( z0 ) m ! f ( z )dz f ( z )dz 1 = − m + 1 ∫ h 2π i c ( z − z0 ) ( z − z0 ) m+1 h
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=
m! 1 1 − f ( z) dz m + 1 m + 1 ∫ 2π ih c ( z − z0 ) ( z − z0 − h )
=
m! f ( z )dz ( z − z0 )m +1 − 1 m + 1 m + 1 ∫ 2π ih c ( z − z0 ) ( z − z0 − h)
= = = = =
m! f ( z )dz ( z − z0 − h) − ( m+1) − 1 2π ih ∫c ( z − z0 )m +1 ( z − z0 )− ( m +1) − ( m +1) m! f ( z )dz z − z0 − h − 1 m + 1 2π ih ∫c ( z − z0 ) z − z0 − ( m +1) m! f ( z )dz h 1 − − 1 m + 1 ∫ 2π ih c ( z − z0 ) z − z0 2 m! f ( z )dz h (m + 1)(m + 2) h 1 + ( m + 1) + + .... − 1 2π ih ∫c ( z − z0 )m +1 z − z0 2! z − z0 m! f ( z )dz h h(m + 1)(m + 2) (m + 1) + + ....... m + 1 ∫ 2π ih c ( z − z0 ) ( z − z0 ) ( z − z0 )2!
Taking limit h→0, we have,
lim h →0
f m ( z0 + h ) − f m ( z 0 ) m ! f ( z ) dz = [(m + 1) + 0 + 0 + ...] ∫ c h 2π i ( z − z0 )m + 2
or , f ( m+1) ( z0 ) =
(m + 1)! f (z) dz ∫ 2π i c ( z − z0 ) m+ 2
Which is true for n=m+1. So by induction hypothesis the theorem is true for all n. Hence, f n ( z0 ) =
n! f ( z ) dz . ∫ c 2π i ( z − z0 ) n +1
Cauchy Inequality: If f(z) is analytic within and on a circle of radius R centre at z0 then f n ( z0 ) ≤ where ‘m’ is constant so that f ( z ) ≤ m on c. Proof: We have,
f n ( z0 ) =
n! f ( z ) dz ∫ 2π i c ( z − z0 ) n +1
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mn ! (for n=0,1,2…) Rn
Then, f n ( z0 ) =
dz n! f ( z ) dz n! ≤ m∫ n +1 ∫ c c 2π i ( z − z0 ) 2π z − z0
n +1
Taking the integral into polar form. Since z-z0=Reiθ, dz=Rieiθdθ. Then, 2π
2π
iθ iθ 2π Rdθ n ! ∫0 Rie dθ n ! ∫0 R i e dθ n! n f ( z0 ) ≤ m = m = m∫ n + 1 n + 1 n + 1 2π 2π 2π 0 R n +1 R eiθ Reiθ
=
n! mn ! 1 m n 2π = n 2π R R Hence , f n ( z0 ) ≤
mn ! . Rn
Morera’s Theorem (Converse of Cauchy): If a function f is continuous throughout a domain D and if
∫
c
f ( z )dz = 0 for every closed
contour c lying in D, then f is analytic throughout D. Proof: Let z0 is a fixed point and z be arbitrary point within simply connected region joining z0 to z. Let c be any closed curve consisting c1 and –c2, then by consequence of Cauchy Gaursat’s theorem we get,
∫
c1
z
f ( z )dz = ∫ f ( z )dz then we can write F(z)= ∫ f ( s)ds....(1) where ‘s’ is the variable of z0
c2
integration lies on c. The integral (1) is independent of the curve joining z0 and z and depends only on z. Let z+∆z be any point distinct from z in R. Then, F(z+∆z)=
Now,
⇒
∫
z + z0
z0
f ( s )ds.......(2)
z F ( z + ∆z ) − F ( z ) 1 z +∆z 1 z +∆z = f ( s )ds − ∫ f ( s )ds = f ( s )ds ∫ z0 ∆z ∫z ∆z ∆z z0
z +∆z F ( z + ∆z ) − F ( z ) 1 z +∆z ∆z 1 z +∆z − f (z) = f ( s )ds − f ( z ). = f ( s )ds − f ( z ) ∫ ds ∫ ∫ z ∆z ∆z z ∆z ∆z z
=
1 z +∆z [ f ( s) − f ( z )] ds ∆z ∫z
But f(s) is continuous at all points z. Hence for each ε>0 ∃ δ>0 s.t. f ( s ) − f ( z ) < ε whenever
s − z < δ. Consequently if the point z+∆z is closed enough to z so that s − z < δ .
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Then,
F (z + ∆ z) − F ( z ) 1 z +∆z 1 − f ( z) = [ f ( s) − f ( z )] ds ≤ ∫ z ∆z ∆z ∆z
∫
z +∆z
z
f ( s ) − f ( z ) ds <
1 ε ∆z = ε ∆z
F ( z + ∆z ) − F ( z ) = f ( z) ∆z → 0 ∆z
∴ lim
i.e. F’(z)=f(z). Thus we have seen that the derivative of F(z) exist for all values of z in R. Therefore F(z) is analytic in R. Consequently F’(z) i.e. f(z) is also analytic in R because derivative of an analytic function is analytic. Liouville’s Theorem: If f(z) is entire function satisfying inequality f ( z ) ≤ M for all values of z where M is a positive constant then f(z) is constant. Proof: Suppose f(z) is an entire function satisfying the inequality f ( z ) ≤ M for all values of z where M is a positive constant, let z1 and z2 be any two points on the complex plane. Let z1 and z2 be any two distinct points and c is a circle with centre z1 and radius r such that z2 lies with in c. Then by Cauchy integral formula we get,
f ( z1 ) =
1 f ( z ) dz 1 f ( z ) dz and f ( z2 ) = ∫ ∫ c c 2π i z − z1 2π i z − z2
∴ f ( z1 ) − f ( z2 ) =
1 1 1 − f ( z )dz ∫ 2π i c z − z2 z − z1
⇒ f ( z2 ) − f(z1 ) ≤
1 2π
∫
c
z2 − z1 f ( z ) dz ......(1) z − z1 z − z2
Taking r is so large s.t. z2 − z1 <
r , z − z1 = r and 2
z − z2 = z − z1 + z1 − z2 ≥ z − z1 − z2 − z1 > r − Then from (1) we get, f ( z2 ) − f ( z1 ) <
r r = and f ( z ) ≤ M ∀z. 2 2
1 z2 − z1 M 2π r 2 z2 − z1 M = → 0 as r→∞. r 2π r r 2
⇒ f ( z2 ) − f ( z1 ) = 0 ⇒ f ( z2 ) = f ( z1 ) i.e. f(z) is constant.
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.z1
.z2
Maximum Modulus Theorem: If f(z) is analytic within and on a simple closed curve c then f ( z ) reaches its maximum value on c unless f(z) is constant. Proof: Suppose f(z) is analytic within and on a simple closed curve c, so f(z) is continuous for the closed curve c. Therefore, it is bounded and it has its maximum value say M. If possible suppose that
f ( z ) has a maximum value inside c. So take ‘a’ is inside c and f (a) = M . Let c1 be a circle inside c centered at a. Since f(z) is not constant within c, there must be a point inside c1 say ‘b’ s.t.
f (b) < M . So that f (b) = M − ε for suitable ε>0. Since f(z) is continuous at b, so for every ε>0 ∃ δ>0 s.t.
f ( z ) − f (b) <
ε 2
.
whenever 0 ≤ z − b < δ .......(1)
i.e. f ( z ) − f (b) ≤ f ( z ) − f (b) <
⇒ f ( z ) < f (b) +
ε
ε
whenever 0 ≤ z − b < δ .
2
ε
= M − ε + ........(2) for all z satisfying z − b < δ . 2 2
That is for all point z inside a circle c2 with centre b and radius δ. Now suppose b − a = r and c3 is the circle with centre a and radius ‘r’ then let PQ is the common arc in c2 and c3 and let ∡PaQ = α . Hence f ( z ) < M −
ε 2
for the arc PQ and f ( z ) < M for the remaining arc. Also ‘a’ is also within
c3. So by Cauchy integral formula,
1 f ( z )dz ∫ c 2π i 3 z − a
f (a ) =
Put, z − a = reiθ , dz = ireiθ dθ . Then
1 2π f ( a + reiθ ) rieiθ dθ 2π i ∫0 reiθ
f (a ) =
f (a) =
1 2π f (a + reiθ )rieiθ dθ 2π i ∫0 reiθ
1 2π 1 f (a) ≤ 2π
f (a) ≤
f (a) <
∫
2π
0
∫
α
0
f (a + reiθ dθ f (a + reiθ dθ +
1 2π
2π
∫α
f (a + reiθ dθ
1 ε 1 M α αε Mα αε ( M − )α + M (2π − α ) = = − +M − =M− 2π 2 2π 2π 4π 2π 2π 17
i.e. f (a) < M −
αε which is impossible. So f ( z ) cannot have max. value within c. So it must 4π
attain its maximum value on c. Minimum Modulus Theorem If f(z) is analytic within and on a simple closed curve c and f(z)≠0 inside c then f ( z ) reaches its maximum value on c. Proof: Suppose f(z) is analytic within and on a simple closed curve c. Since f(z)≠0 inside c, so is analytic within c, then by maximum modulus theorem
1 f ( z)
1 has its maximum value on c. Since f (z)
1 is maximum iff f ( z ) is minimum. So f ( z ) has its minimum value on c. f (z) Fundamental Theorem of Algebra Any polynomial Pn(z)=a0+a1z+a2z2+…….+anzn, ai∈ℂ of degree n n≥1 with an≠0 has at least one zero. That is, there exist at least one point z0 s.t. Pnz0=0. Proof: Suppose Pnz≠0 for all z∈ℂ. Then
1 1 = = Pn ( z ) a0 + a1 z + a2 z 2 + ... + an z n
1 is entire function. Then it is bounded. Since, Pn ( z ) 1
a a z an + n−1 + ... + 0n z z
→ 0 as z→∞.
n
Then by Liouvilles theorem, any bounded entire function is constant and so Pnz is constant. This is a contradiction. Since Pnz is a polynomial of degree greater than 1. So Pnz must have at least one zero.
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