Deformation Theorem

Lecture 16

Deformation Theorem 103

and

γ_1b

f(z)dz =

P₂P₄

+

γ_2b

+

P₃P₁

f(z)dz.

Figure 16.2 P₁

P₂

γ_2a γ_2b

P₃

P₄ P₁

P₂ P₃

P₄ γ_1a

γ_1b

Adding these equations, we find

γ₁

f(z)dz =

γ_1a

+

γ_1b

f(z)dz

=

γ_2a

+

γ_2b

f(z)dz =

γ₂

f(z)dz.

Example 16.1.

Determine the possible values of

γ

1

(z−a)dz,where γ is any positively oriented, simple, closed contour not passing through z=a.Observe that the function 1/(z−a) is analytic everywhere except at the pointz=a.Ifalies exterior toγ,then the integral is zero by Theorem 15.2. Ifalies insideγ,we choose a small circleγr centered ataand lying within γ.Then, from Theorem 16.1, it follows that

γ

dz z−a =

γ_r

dz z−a.

Now, since, onγr, z =a+re^iθ (ris fixed),dz=rie^iθdθ, and hence

γ_r

dz z−a =

2π 0

1

re^−iθrie^iθdθ = 2π

0

idθ = 2πi.

Hence, we have

γ

dz z−a =

2πi ifais inside γ 0 ifais outside γ.

Example 16.2.

Evaluate

γ

1

z²−1dz,where the contourγ is given in Figure 16.3.

Figure 16.3

·

^γr

·

−1 1

γ

Clearly, the integrand 1/(z²−1) fails to be analytic at z=±1.Since the point 1 lies outside ofγ,the integral overγis the same as the integral over that small circleγrenclosing−1.Using partial fraction expansion, we find

γ

dz z²−1 =

γ_r

dz z²−1 =

γ

1

2(z−1)− 1 2(z+ 1)

dz

= 0−1

22πi = −πi.

Now we state the following result that extends Theorem 16.1.

Theorem 16.2.

Letγ, γ₁,· · ·, γ_nbe simple, closed, positively oriented contours such that eachγ_j, j= 1,· · ·, nlies interior to γ,and the interior of γ_j has no points in common with the interior ofγ_k ifj =k (seeFigure 16.4, where for simplicity we have taken each γ_j to be a circle). If f is analytic on the closed region containing{γ}, {γ₁},· · ·,{γ_n}and the points between them, then

γ

f(z)dz = n j=1

γ_j

f(z)dz. (16.1)

Figure 16.4 γ₁ γ₂ γ

γ₃

γ_n

Deformation Theorem 105

Proof.

The proof of Theorem 16.2 is exactly the same as that of Theorem 16.1, except here we need to divide γinto several parts.

Example 16.3.

Find

γ

(5z−2)

z²−z dz,where γis given in Figure 16.5.

Figure 16.5

0

·

1

·

γ

The integrandf(z) = (5z−2)/(z²−z) is analytic everywhere except for the zeros of the denominator,z= 0 and z= 1.Letγ1andγ2 be two small circles enclosing these points. Then, using Theorem 16.2, we get

Figure 16.6

0

·

1

·

γ γ₁

γ₂

γ

f(z)dz =

γ₁

f(z)dz+

γ₂

f(z)dz.

Thus, we have

γ

5z−2 z(z−1)dz =

γ₁

2 z + 3

z−1

dz+

γ₂

2 z + 3

z−1

dz

= 2×2πi+ 0 + 0 + 3×2πi = 10πi.

Example 16.4.

Let γ be a simple closed contour that contains the distinct pointsz1, z2,· · ·, zn in its interior. We shall show that

γ

dz

(z−z₁)(z−z₂)· · ·(z−z_n) = 0.

For this, we write the partial fractional decomposition 1

(z−z₁)(z−z₂)· · ·(z−zn) = A1

z−z₁+ A2

z−z₂ +· · ·+ An

z−zn

,

which is the same as 1 =

n j=1

Aj(z−z1)· · ·(z−zj−1)(z−zj+1)· · ·(z−zn).

Thus, comparing the coefficients of zⁿ⁻¹ on both the sides, we find that A1 +A2 +· · · +An = 0. Next, let γj, j = 1,2,· · ·, n be small circles with centerzj, that lie in the interior ofγ.Then, from Theorem 16.2 and Example 16.1, it follows that

γ

dz

(z−z₁)(z−z₂)· · ·(z−zn) = n j=1

γ_j

Aj

z−zj

= 2πi n j=1

Aj = 0.

Problems

16.1. Use an antiderivative to show that for everyγ extending from a pointz1 to a pointz2

γ

zⁿdz = 1 n+ 1

z₂ⁿ⁺¹−zⁿ⁺¹₁

for n= 0,1,2,· · ·.

16.2. Letγbe the semicircle from−3ito 3iin anticlockwise direction.

Show that

γ

dz z =πi.

16.3. Evaluate each of the following integrals where the path is an arbitrary contour between the limits of integrations

(a).

i/2 i

e^πzdz, (b).

π+2i 0

cos z

2

dz, (c).

3 1

(z−3)³dz.

16.4. Show that ifzⁱ is the principal value, then

γ

zⁱdz = 1 +e^−π 2 (1−i), whereγ is the upper semicircle fromz= 1 toz=−1.

16.5. Let f and g be analytic for all z, and let γ be any contour joining the pointsz₁ andz₂.Show that

γ

f(z)g(z)dz = f(z₂)g(z₂)−f(z₁)g(z₁)−

γ

f(z)g(z)dz.

Deformation Theorem 107 16.6. Let f = u+iv be continuous and possess continuous partial derivatives with respect toxandy in a neighborhood ofz= 0.Show that f is differentiable atz= 0 if and only if

rlim→0

1 πr²

|z|=r

f(z)dz = 0. (16.2)

16.7. Use Green’s Theorem to show that Area of D = − i

2

C

zdz.

16.8. For each of the following functionsf,describe the domain of ana- lyticity and apply the Cauchy-Goursat Theorem to show that

γ

f(z)dz= 0,whereγis the circle|z|= 1 :

(a).f(z) = 1

z²+ 2z+ 2, (b).f(z) =ze^−z. What about (c).f(z) = (2z−i)⁻²?

16.9. Let γ denote the boundary of the domain between the circle

|z| = 4 and the square whose sides lie along the lines x = ±1, y = ±1.

Assuming thatγis oriented so that the points of the domain lie to the left ofγ,state why

γ

f(z)dz= 0 when (a).f(z) = z+ 2

sinz/2, (b).f(z) = z 1−e^z.

16.10. Let γ denote the boundary of the domain between the circles

|z| = 1 and |z| = 2. Assuming that γ is oriented so that the points of the domain lie to the left of γ, show that

γ

f(z)dz = 0 for the following functions:

(a).f(z) = e^z

z²+ 9, (b).f(z) = cotz.

16.11.Letγ be the unit circle |z|= 1 traversed twice in the clockwise direction. Evaluate

(a).

γ

Log (z+ 2)dz, (b).

γ

dz 3z²+ 1.

16.12.Let r, R be constants such that 0 < r < R. Denote by γr the circlez=re^iθ, 0≤θ≤2π.Show that

1 2πi

γ_r

R+z

(R−z)zdz = 1.

Hence, deduce that 1 2π

2π 0

R²−r²

R²+r²−2rRcosθdθ = 1.

16.13.Suppose that f is of the form f(z) =

n j=1

Aj

z^j +g(z),

where g is analytic inside and on the simple, closed, positively oriented contourγcontaining 0 in its interior. Show that

γ

f(z)dz= 2πiA₁. 16.14.Establish the following Wallis formulas

(a).

2π 0

cos^2ktdt = 2π (2k)!

2^2k(k!)², (b).

2π 0

cos^2k+1tdt = 0, (c).

2π 0

sin^2ktdt = 2π (2k)!

2^2k(k!)², (d).

2π 0

sin^2k+1tdt = 0.

Answers or Hints

16.1. !

γzⁿdz= _n+1¹ (zⁿ⁺¹₂ −z₁ⁿ⁺¹).

16.2. !

γdz/z= Logz³ⁱ

−3i (since _dz^dLogz = ¹_z on γ) = ln|3|+i^π₂ − ln|3| −^iπ₂

=iπ.

16.3. (a). (1 +i)/π,(b).e+e⁻¹,(c). −4.

16.4. Principal value zⁱ =eⁱLogz. Hence, !

γzⁱdz =−!0

πe^i(iθ)ie^iθdθ =

−⁽¹⁻₂ⁱ⁾(e^−π+ 1).

16.5. Use [f(z)g(z)]=f(z)g(z) +f(z)g(z).

16.6. As in Lecture 15, we have

!

|z|=rf(z)dz=! !

|z|≤r(−v_x−u_y)dxdy+i! !

|z|≤r(u_x−v_y)dxdy, and hence, from the Mean Value Theorem of integral calculus, we find

1 πr²

!

|z|=rf(z)dz=−(vx(x, y) +uy(x, y)) +i(ux(ˆx,y)ˆ −vy(ˆx,ˆy)), where (x, y) and (ˆx,y) are suitable points. Thus, (16.2) holds if and only ifˆ the Cauchy-Riemann conditions (6.5) hold atz= 0.Now, since the partial derivatives ofuandvare continuous, it is equivalent to the differentiability off atz= 0.

Deformation Theorem 109 16.7. In Green’s Theorem, we letP =xandQ=y,to obtain 2! !

Ddxdy=

!

Cxdy−ydx, and hence 2D = !

Cxdy−ydx. Suppose the parametric equation of C is x = x(t), y = y(t), so that z(t) = x(t) +iy(t) and x(a) = x(b), y(a) = y(b) (C is closed). Then 2D = !b

a(xy −yx)dt.

Now !

Czdz =!b

a(x−iy)(x+iy)dt =!b

a[(xx+yy) +i(xy−yx)]dt=

12[x²(t) +y²(t)]^b

a+i!b

a(xy −yx)dt = 0 +i!b

a(xy −yx)dt = 2iD, and henceD=_2i¹ !

Czdz=−ⁱ₂!

Czdz.

16.8. (a). f(z) = _{(z−(−1+i))(z}¹ _{−(−1−i))} and thus analytic if z=−1 +i or

−1−i; furthermore,!

γ dz

z²+2z+2 = 0, (b). ze^−z is entire. (c). Analytic if z =i/2,cannot apply Cauchy-Goursat Theorem; however, sincef has an antiderivative onγ, !

γ

(2zdz−i)² = 0.

16.9. (a).f(z) = [(sinz/2)(1)−(z+2)(1/2) cosz/2]/sin²z/2 if sinz/2= 0;

sinz/2 = 0 if and only if e^−iz/2(e^iz−1) = 0; i.e., if and only if e^iz = 1, or z = 2kπ, k ∈ Z; none of this point is in the domain. (b). f(z) = [(1−e^z)−z(−e^z)]/(1−e^z)²ife^z= 1;e^z= 1 if and only ifz= 2kπi, k∈Z;

none of this point is in the domain. Finally, note that γ = γ₁+γ₂ and

!

γ₁f dz=!

γ₂f dz= 0.

γ₁

γ₂

16.10.(a). f(z) =e^z/(z²+ 9) is analytic ifz²+ 9= 0; i.e.,z=±3i.But

±3i are not in the domain, and hence, by the Cauchy-Goursat Theorem,

!

γ₁f(z)dz=!

γ₂f(z)dz= 0.Hence,!

γf(z)dz= 0.

γ γ₁

γ₂

(b). cotzis analytic if sinz= 0; i.e.,z=kπ, k∈Z.Butkπ are not in the domain for allk∈Z.Hence, as in (a),!

γcotzdz= 0.

16.11. Let γ be the circle |z| = 1 taken in the anticlockwise direction.

(a).!

γLog (z+ 2)dz= 0 by the Cauchy-Goursat Theorem, (b).!

γ

3zdz²+1 =

13

!

γ

_√

3

2i(^z−i/√3)−_2i(^z+i/√3√3)

dz= ^√_6i³(2πi)−^√_6i³(−2πi) = 0.Thus,

−!

γLog(z+ 2)dz=!

γLog(z+ 2)dz+!

γLog(z+ 2)dz= 0 and−!

γ

3zdz²+1 =

!

γ

3zdz²+1+!

γ

3zdz²+1 = 0.

16.12.Use partial fractions. Compare the real parts.

16.13.Use Theorem 16.1 and Examples 14.6 and 16.1.

16.14. Let γ be the unit circle |z| = 1. Then, _2πi¹ !

γ

z+¹_{zn dz}

z = ²_2πⁿ×

!2π

0 cosⁿtdt. Now expand (z+ 1/z)ⁿ using the binomial formula, and use Problem 16.13.

Lecture 17