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AE 525a/ME 525: Lecture #8

0.1 Analytic functions

Definition 1 A function f(z) of a complex variable z is said to have a deriva-tive if

limz

f() f(z) z

exists and has the same value regardless of the mode in which z. Then thislimit is denoted by f0(z) or df/dz.

Different modes of z.

Example 2

f(z) = z3

f0(z) = limz

3 z3 z = limz(

2 + z + z2) = 3z2

Notation 3 Formally the definition is the same as in the case of a real variablehowever, independence of the way in which z is very restrictive.

Example 4

f(z) = zz

limz

zz z = limz z + z zz z

= limz

(+ z z z )

= limz

(+ zei2)

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which is different for different . Therefore, f0(z) does not exist, except at theorigin.

The value of

z = ei as

z.

Definition 5 If f(z) has derivative at a point z0 and in some neighborhood ofz0, then f(z) is said to be analytic at z0 (holomorphic, regular).

Since the definition of the derivative f0(z) is formally the same as for the realvariable functions, the rules of differentiation of composite expressions knownfor real domain variable must hold for complex variable as well, i.e.,

(af(z) + bg(z))0 = af0(z) + bg0(z)

(f g)0 = f0g + f g0

(f /g)0 =f0g f g0

g2; g 6= 0

[f(g(z))]

0

= f

0

(g)g

0

(z)Example 6

dez

dz= lim

z

e ez z = limz e

z ez 1 z

= ez limz

1 + ( z) + (z)22! + ... z

= ez limz

[1 + z

2!+ ...] = ez

Example 7 Similarly, we can show that

d

dz sin z = limz

sin(

z + z)

sin(z)

z = cos zd

dzln z =

1

zd

dztan1 z =

1

1 + z2, etc.

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0.2 Branch Cuts and Riemann Surfaces

Definition 8 Riemann surface is a generalization of the zplane to a surfaceof more then one sheet such that a multiple-valued function has only one valuecorresponding to each point on that sheet.

Example 9 Consider the function

f(z) = z1/2 = w

z = rei(+2n); n = 0, 1, 2,... w =

rei(/2+n)

n = 0 w = rei/2 = w1n = 1 w = rei/2 = w2n = 3 w = rei/2 = w1, etc.

Therefore, there will be two possible values of the functionw1 andw2, called thebranches of w as increases in multiples of 2.

Increase of the argument in increments of 2.

Therefore, for the function f(z) = z1/2 there are two types of paths:

Path 1, where never increases for 2, and

path 2, where increases beyond 2 resulting in double values w1 and w2.

Thus the origin possesses the property that a loop about it interchanges

the branches and thus the origin O is defined as branch point of the function.

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Different paths for the function z1/2.

To prevent the multiple valuedness we introduce a branch cut. For exam-ple, for w = z1/2 the branch cut is depicted by the following figure

A branch cut for the function z1/2.

If we choose the value of w at z0 to be, say, w1(z0) and require that thevalue of w at any other point z1 be obtained by continuous variation of walong any curve joining z0 and z1, then the fact that loops about the origin areprevented means that z1/2 is uniquely defined.

Example 10 The function f(z) =p

(z a)(z b), where a and b are real hasbranch points at z = a and z = b. Lets define the branch cuts as

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Branch cuts forp

(z a)(z b).

which implies that

z a = r1ei1 ; < z b = r2ei2 ;

2 < 3

2

Then the functionf(z) =

r1r2e

i(1/2+2/2)

is uniquely defined in the complex plane.

Example 11 Consider the function f(z) = (z a)1/n1(z b)1/n2 with branchpoints at z = a and z = b. Using

z a = r1ei(1+2n)

z b = r2ei(2+2n)

we have that

f(z) = r1/n11 r

1/n22 e

i[1/n1+2/n2+2n(1/n1+1/n2)]

If1

n1+

1

n2= m

where is an integer, then thefinite branch cut is possible and the function issingle valued in the complex plane as shown.

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Finite branch cut for the function (z a)1/n1(z b)1/n2 when1/n1 + 1/n2 = m(an integer).

Example 12 Consider the functionf(z) = ln[(z a)(z b)]

= ln(r1r2) + i(1 + 2 + 4n)

where

z a = r1ei(1+2n)z b = r2ei(2+2n)

which has the branch points at z = a and z = b. Obviously, the finite branchcut is unacceptable since we are not allowed to increase 1 or 2 for 2 whichwould introduce multiple-valuedness of f. Thus the following branch cuts areacceptable

Branch cuts for the function ln[(z a)(z b)].Example 13 Consider the function

f(z) = lnz

a

z b = lnr1r2 + i(

1 2)where

z a = r1ei(1+2n)z b = r2ei(2+2n)

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has the branch points at z = a and z = b. Therefore, the finite branch cut isOK.

Example 14 For the function

f(z) = zp

q = (z1

q )p

has a branch point atO and a semiinfinite branch cut ai acceptable. For example,

f(z) = z1

n

where n is an integer can be represented as

f(z) = r1

n ei(/n+2m/n)

and the branch cut can be chosen as

Branch cut for f(z) = z1/n.

Notation 15 We should note that both the function and all its derivativesf(z) = z1/n have algebraic branch point at O. However, the functionf(z) = ln zhas a logarithmic branch point at O while its derivatives do not have logarithmicbranch point.

Example 16 Lets reconsider the function

f(z) =p

(z a)(z b) = r1/21 r1/22 ei[1/2+2/2+n+m]

as the complex variable z varies along various closed loops in the zplane:

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Different paths for the functionp

(z a)(z b).Thus we have

path 1 : 2 2 + 2; 1 1 f(z) f(z)path 2 : 2 2; 1 1 + 2 f(z) f(z)path 3 : 2 2 + 2; 1 1 + 2 f(z) f(z)

Therefore, thefinite branch cut is OK.

0.3 Cauchy-Riemann Equations

Consider an analytic function f(z) = u(x, y) + i(v(x, y). Then,

f0(z0) = limzz0

f(z) f(z0)z z0

= limzz0

[u(x, y) u(x0, y0)] i[v(x, y) v(x0,y0)](x x0) + i(y y0)

Since the mode of z approaching z0 arbitrary, we choose first that along theline y0, i.e., from point (x0, y0) to (x, y0). Therefore,

f0(z0) = limzz0

u(x, y) u(x0, y0)

(x x0)+ i

v(x, y) v(x0,y0)(x x0)

=u(x0, y0)

x+ i

v(x0, y0)

x

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Similarly, by taking the path from (x0, y0) to (x0, y) we have that

f0(z0) = limzz0

u(x0, y) u(x0, y0)

i(y y0)+ i v(x0, y) v(x0,y0)

i(y y0)

= iu(x0, y0)y

+v(x0, y0)

y

Therefore, we get the Cauchy-Riemann equations

u(x, y)

x=

v(x, y)

y

u(x, y)

y= v(x, y)

x

which means that if the derivative f0(z) exists, then the Cauchy-Riemann equa-

tions are satisfied. Thus the Cauchy-Riemann equations are necessary conditionfor existence of derivative f0(z).

Proposition 17 The Cauchy-Riemann equations are sufficient condition forexistence of a derivative.

Proof. Sufficient condition for existence of derivative f0(z) (i.e., if ux =vy&uy = vx f0(z) exists). From Taylor series we have that

u(x, y) u(x0, y0) = ux(x0, y0)(x x0) + uy(x0, y0)(y y0) + ...v(x, y) v(x0,y0) = vx(x0, y0)(x x0) + vy(x0, y0)(y y0) + ...

and consequently

limzz0

u(x, y) u(x0, y0) + i[v(x, y) v(x0,y0)]

(x x0) + i(y y0)

= limzz0

ux(x0, y0)(x x0) + uy(x0, y0)(y y0) + ...

(x x0) + i(y y0)

+i limzz0

vx(x0, y0)(x x0) + vy(x0, y0)(y y0) + ...

(x x0) + i(y y0)

= limzz0

ux(x0, y0)(x x0) vx(x0, y0)(y y0) + ...

(x x0) + i(y y0)

+ limzz0

i

vx(x0, y0)(x x0) + ux(x0, y0)(y y0) + ...

(x x0) + i(y y0)

= ux(x0, y0) + ivx(x0, y0) = f

0

(z0)

Thus we have proved that Cauchy-Riemann equations are necessary andsufficient condition for analyticity.

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Notation 18 We should note that from the Cauchy-Riemann equations it fol-lows that

ux = vy&uy = vx uxx + uyy = 0 = 2uvxx + vyy = 0 = 2v

or that real and imaginary parts of an analytical function are harmonic.

Example 19 No purely real function can be analytic unless it is a constant.Namely, let Im(f(z) = v(x, y) = 0 or

f(z) = u(x, y)

If f(z) is analytic it must satisfy the Cauchy-Riemann equations, thus

ux = vy = 0 u(x) = g(y)uy = vx = 0 g0(y) = 0 g(y) = const

sou(x) = g(y) = const

0.4 Integration in the Complex Plane

Consider a path C in a plane of a complex variable z. Let f(z) be any complexfunction (analytic or not). Then

ZCf(z)dz limn;|zj |0

n

Xj=1 f(j)

zj (1)

this limit is called integral of f(z) along C, provided that the limit does notdepend on the way in which zj and j are chosen. In terms of x and y we havethat Z

C

f(z)dz =

ZC

udx vdy + iZC

f dx + udy

so that for these integrals the real variable theory applies.Immediate consequence of definition (1) is that the usual rules for manipu-

lation of integrals apply:

ZC:AB

f(z)dz = ZC:BA

f(z)dz

where C : AB denotes moving along the curve C from points A to B. Similarly

ZC

[af(z) + bg(z)]dz = a

ZC

f(z)dz + b

ZC

g(z)dz, etc.

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Notation 20 Iff(z) is bounded onC, i.e., | f(z) | M, z C, then the triangleinequality can be applied to (1) to obtain the following result

|

ZC

f(z)dz |ZC

|f(z)||dz| ML

where L denotes the length along the curve of integration C.

Theorem 21 Cauchy Theorem. Letf(z) be an analytic function in R, and letC denotes a closed curve completely contained in R. Then

IC

f(z)dz = 0 (2)

where unless stated differently the integration along C takes place in counter-clockwise direction.

Integration path in Cauchy theorem.

Proof. Recall first the application of divergence theorem to a region Abounded by the closed curve C

ZA

divudA =

IC

u nds

for a vector-valued function u = (u, v) results in

ZA

(ux + vy)dA =

IC

(unx + vny)ds

where n denotes an outward unit normal to C and s denotes the length of thecurve. It is easy to show that

dsny = dxdsnx = dy

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so that

ZA

(ux+vy)dA =IC

(unx+vny)ds =ZA

(ux+vy)dA =IC

(udyvdx) Green0s theorem

Therefore ZA

(ux vy)dA =IC

(udy + vdx)

Consequently

IC

f(z)dz =

IC

(u + i v)(dx + i dy)

=

IC

(udx vdy) + iIC

(vdx + udy)

= IC(vx + uy)dA + i IC(ux vy)dA = 0

Notation 22 The proof Greens theorem requires continuity of partial deriva-tives ofu andv (i.e., f0(z) continuous). Goursat proved Cauchy theorem underless restricted conditions.

0.5 Multiple Connected Regions

Definition 23 A simply connected region is defined as the region in which aclosed curve drawn in the region encloses only the points belonging to the region.

For a region R not simply connected and for f(z) analytic and single valuedin R we have that I

C1

f(z)dz 6= 0

Namely, by introducing appropriate cuts, the region becomes simply con-nected. Therefore we get that the so called generalization of Cauchy The-orem I

C1

f(z)dz IC2

f(z)dz IC3

f(z)dz = 0

where the integration along closed loops takes place in the counterclockwisedirection.

Example 24 Show that IC

dzz

= 2i

for any closed curve C enclosing the origin.

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Figure 1: Multiple connected region R.

Integration contour forHCdz/z.

First we isolate the singularity by excluding a small circle C of radius from the domain. Therefore, with appropriate cuts the new domain is simplyconnected and the function 1/z is single valued. Thus the generalize CauchyTheorem results in

IC

dz

z= lim

0

IC

dz

z= lim

0

Z20

iei

eid = 2i

0.6 Cauchys Integral Formula

Cauchys integral formula expresses the value of a function f(z) at any pointinside a contour in terms of its values on the contour.

Theorem 25 Cauchys Integral Formula. If f(z) is analytic in a simply con-

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nected region R and if C is any closed contour in R, then

f(z) = 12iIC

f() z d

Proof. Lets isolate the singularity of the integrand at = z by removingan infinitesimal circle C centered at z.Then the integrand becomes analyticand single valued everywhere inside the contour C+ C + l+ + l. Consequently,Cauchy theorem implies thatI

C

f()

z

d = lim0

IC

f()

z

d

= lim0

IC

f(z) + f() f(z) z d

= lim0

IC

f(z)

z d+ lim0IC

f() f(z) z d

But using the result from the previous example we have that

lim0

IC

f(z)

z = f(z)2i

We shall prove later that iff(z) has derivative at z = z0 then it is also continuousat z0, i.e., | f(z) f(z0) |< for z z0. Thus as 0

IC |

f()

f(z)

z || d | Z2

0

d = 2 0as 0. Therefore, I

C

f()

z d= 2if(z)

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Notation 26 If z is outside the contour C,then F()/( z) is analytic in Rand thus I

C

f() z d= 0

0.6.1 Case of Multiple Connected Regions

Lets consider the multiple connected domain as shown bellow.

Then with appropriate cuts, the domain becomes simply connected and theCauchy integral formula becomes

1

2i

IC

f()

z d1

2i

IC1

f()

z d1

2i

IC2

f()

z d

=1

2ilim0

IC

f()

z d= f(z)

So that the corresponding Cauchy integral formula becomes

f(z) =1

2i

IC

f()

z d1

2i

IC1

f()

z d1

2i

IC2

f()

z d

0.7 Integrals Dependent on a ParameterRecall the Cauchys integral formula

f(z) =1

2i

Ic

f()

z d

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where is a variable of integration and z is a parameter. In general, we consider

F(z) =IC(z, )d (3)

z = x + iy; = + i

and C is a closed piecewise smooth curve.

The function is the function of two complex variables satisfying the fol-lowing properties:

1. (z, ) is analytic for z D and C.2. and /z are continuous functions of z, for z D and C.

Claim 27Then the function F(z) is an analytic function of the complex vari-ablez inD and derivative of F(z) may be computed by differentiating under the

integral sign.

Proof. Consider

F(z) =

IC

(z, )d=

IC

(u + iv)(d+ id)

F(z) = U(x, y) + iV(x, y) =

IC

(ud v) + iIC

(vd+ ud)

U(x, y) =

IC

(ud v); V(x, y) =IC

(vd+ ud)

Now the functions u and v possess partial derivatives w.r. to x and y and arecontinuous in x and y. Thus, the partial derivatives w.r. to x and y of U existsand may be computed by differentiating under the integral sign. Then

Ux =

IC

(uxd vx)

Uy =

IC

(uyd vy)

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are continuous functions of x, y in D.Similarly,

Vy =IC

(vyd+ uyd) =IC

(uxd vxd) = Ux

Vx =

IC

(vxd+ uxd) =

IC

(uyd+ vyd) = Uy

or the functions U and V satisfy the Cauchy-Riemann equations. Consequently,the function F(z) = U + iV is analytic in D. Note that

F0(z) = Ux + iVx =

IC

(uxd vx) + iIC

(vxd+ uxd)

=

IC

(ux + ivx)(d+ i) =

IC

zd

Thus to compute the derivative of an integral F(z) =HC(z, )d, (when is

analytic for C and z D, and and /z are continuous functions of and z) one differentiates the integrand w.r to parameter z.

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