Complex Integration

Lecture 13

and b

a

z0w1(t)dt = b

a

[x0u1(t)−y0v1(t)]dt+i b

a

[x0v1(t) +y0u1(t)]dt

=

x0

b a

u1(t)dt−y0

b a

v1(t)dt

+i

x0

b a

v1(t)dt+y0

b a

u1(t)dt

= (x0+iy0) b

a

u1(t)dt+i b

a

v1(t)dt

= z0

b a

w1(t)dt.

Example 13.1.

Supposew(t) is continuous on [a, b] andw(t) exists on (a, b).For such functions, the Mean Value Theorem for derivatives no longer applies; i.e., it is not necessarily true that there is a numberc∈(a, b) such thatw(c) = (w(b)−w(a))/(b−a).To show this, we consider the function w(t) = e^it, 0 ≤ t ≤ 2π. Clearly, |w(t)| = |ie^it| = 1, and hence w(t) is never zero, whilew(2π)−w(0) = 0.

We recall that if f : [a, b] → IR, then |!b

af(x)dx| ≤ !b

a|f(x)|dx. We shall now show that the same inequality holds forw: [a, b]→C; i.e.,

b a

w(t)dt ≤

b a

|w(t)|dt, a≤b <∞.

For this, let |!b

aw(t)dt| =r, so that !b

a w(t)dt=re^iθ in polar form. Now we have

r = e^−iθ b

a

w(t)dt = b

a

e^−iθw(t)dt

= b

a

Re

e^−iθw(t) dt+i

b a

Im

e^−iθw(t) dt (= 0 since LHS is real)

= b

a

Re

e^−iθw(t) dt

≤

b

a

Re

e^−iθw(t) dt

≤

b a

Re

e^−iθw(t)dt

≤ b

a

e^−iθw(t)dt = b

a

|w(t)|dt (since |e^−iθ|= 1).

Complex Integration 85 We note that the Fundamental Theorem of Calculus also holds: If W(t) = U(t) +iV(t), w(t) = u(t) +iv(t) and W(t) = w(t), t ∈ [a, b],

then b

a

w(t)dt = W(b)−W(a).

Thus, in particular, we have b

a

e^z0tdt = 1 z₀e^z0t

^b

a

= 1 z₀

e^z0b−z^z0a

, z₀= 0.

Thelengthof a smooth curveγ given by the range of z : [a, b]→C is defined by

L(γ) = b

a

|z(t)|dt.

Example 13.2.

Forz(t) =z₁+t(z₂−z₁), 0≤t≤1,we have z(t) = z₂−z₁, and hence L(γ) =

1

0 |z₂−z₁|dt = |z₂−z₁|.

Example 13.3.

Forz(t) =z₀+re^it, 0≤t≤2π,we have z(t) = ire^it, and hence L(γ) =

2π

0 |ire^it|dt = 2πr.

Now letS be an open set, and letγ,given by the range ofz: [a, b]→C, be a smooth curve inS. Iff :S→Cis continuous, then the integral off along γis defined by

γ

f(z)dz = b

a

f(z(t))z(t)dt.

Figure 13.1

·

^γ

·

S

The following properties of the integration above are immediate:

γ

[f(z)±g(z)]dz =

γ

f(z)dz±

γ

g(z)dz,

γ

z₀f(z)dz = z₀

γ

f(z)dz, z₀∈C,

−γ

f(z)dz = −

γ

f(z)dz.

Example 13.4.

Let f(z) = z−1 and let γ be the curve given by z(t) =t+it², 0≤t≤1.Clearly,γ is smooth and

γ

f(z)dz = 1

0

f(z(t))z(t)dt = 1

0

(t+it²−1)(1 + 2it)dt

= 1

0

(t−1−2t³)dt+i 1

0

(2t(t−1) +t²)dt, which can now be easily evaluated.

Example 13.5.

Let f(z) = z+ (1/z) for z = 0 andγ be the upper semi-circle at the origin of radius 1; i.e., z(t) =e^πit, 0≤t≤1.Clearly, γ is smooth and

γ

f(z)dz = 1

0

πi

e^πit+e^−πit

e^πitdt = πi.

Thelengthof a contourγ=γ₁+· · ·+γn is defined by L(γ) =

n j=1

(length ofγj) = n j=1

L(γj).

Iff :S→Cis continuous on{γ},then thecontour integral of f along γ is defined by

γ

f(z)dz = n j=1

γ_j

f(z)dz.

Example 13.6.

Letf(z) =z−1 andγ=γ₁+γ₂,where γ₁ is given byz₁(t) =t, 0≤t≤1 andγ₂ is given by z₂(t) = 1 +i(t−1), 1≤t≤2.

Clearly, the contourγ is piecewise smooth, and

γ

f(z)dz = 1

0

f(z1(t))z₁(t)dt+ 2

1

f(z2(t))z₂(t)dt

= 1

0

(t−1)dt+ 2

1

i(t−1)idt, which can be evaluated.

Theorem 13.1 ( ML -Inequality).

Suppose thatf is continuous on an open set containing a contour γ, and |f(z)| ≤ M for all z ∈ {γ}. Then, the following inequality holds

γ

f(z)dz

≤ M L,

Complex Integration 87 whereLis the length of γ.

Proof.

First assume that γ given by the range of z : [a, b] → C is a smooth curve. Then, we have

γ

f(z)dz =

b a

f(z(t))z(t)dt ≤

b a

|f(z(t))||z(t)|dt

≤ M

b a

|z(t)|dt = M L.

Ifγ=γ₁+γ₂+· · ·+γ_n, whereγ₁, γ₂,· · ·, γ_n are smooth, then we find

γ

f(z)dz =

n j=1

γ_j

f(z)dz ≤

n j=1

γ_j

f(z)dz

≤ n j=1

M L(γ_j) = M L(γ).

Example 13.7.

Letγbe given by z(t) = 2e^it, 0≤t≤2π.Show that

γ

e^z z²+ 1dz

≤ 4πe² 3 .

Since|e^z|=e^x≤e²and|z²+ 1| ≥ ||z²| −1|=|4−1|= 3,it follows that

γ

e^z z²+ 1dz

≤ e²

3 ×2×2π = 4πe² 3 .

Problems

13.1. Evaluate the following integrals (a).

1 0

(1 +it²)dt, (b).

π/4

−π

te^−it2dt, (c).

π 0

(sin 2t+icos 2t)dt, (d).

2 1

Log(1 +it)dt.

13.2. Find the length of the arch of the cycloid given byz(t) =a(t− sint) +a i(1−cost), 0≤t≤2πwhereais a positive real number.

13.3. Letγbe the curve given byz(t) =t+it², 0≤t≤2π.Evaluate

γ

|z|²dz.

13.4. Letf(z) =y−x−3ix²andγbe given by the line segmentz= 0 toz= 1 +i.Evaluate

γ

f(z)dz.

13.5. Letγ be given by the semicirclez = 2e^iθ, 0≤θ≤π. Evaluate

γ

z−2 z dzand

−γ

|z^1/2|exp(iArgz)dz.

13.6. Show that ifm andn are integers, then 2π

0

e^imθe^−inθdθ =

0 when m=n 2π when m=n.

Hence, evaluate

γ

z^mzⁿdz, where γ is the circle given by z = cost + isint, 0≤t≤2π.

13.7. Letγbe the positively oriented ellipsex² a²+y²

b² = 1 witha²−b²=

1.Show that

γ

√ dz

1−z² = ±2π, where a continuous branch of the integrand is chosen.

13.8. Letz1(t), a≤t ≤b andz2(t), c≤t≤dbe equivalent param- eterizations of the same curve γ; i.e., there is an increasing continuously differentiable function φ : [c, d] → [a, b] such that φ(c) = a and φ(d) = b and z₂(t) =z₁◦φ(t) for all t ∈[c, d]. Show that if f is continuous on an open set containingγ,then

b a

f(z₁(t))z₁(t)dt = d

c

f(z₂(t))z₂(t)dt.

13.9. Letγ be the arc of the circle|z|= 2 fromz = 2 toz= 2i that lies in the first quadrant. Without evaluating the integral, show that

γ

dz z²−1

≤ π 3. 13.10.Letγ be the circle|z|= 2.Show that

γ

1 z²−1dz

≤ 4π 3 .

13.11.Show that if γ is the boundary of the triangle with vertices at z= 0, z = 3i,andz=−4 oriented in the counterclockwise direction, then

γ

(e^z−z)dz

≤ 48.

Complex Integration 89 13.12.Let γ_R be the circle|z| =R described in the counterclockwise direction, whereR >0.Suppose Logz is the principal branch of the loga- rithm function. Show that

γ_R

Logz z² dz

≤ 2π

π+ LogR R

.

13.13.Let γR be the circle|z| =R described in the counterclockwise direction, whereR >2.Suppose Logz is the principal branch of the loga- rithm function. Show that

γ_R

Logz² z²+z+ 1dz

≤ 2πR

π+ 2LogR R²−R−1

.

13.14. Let S be an open connected set and γ a closed curve in S.

Supposef(z) is analytic onS and the derivativef(z) is continuous onS.

Show that

I=

γ

f(z)f(z)dz is purely imaginary.

13.15. Let f(z) be a continuous function in the region |z| ≥ 1, and suppose the limit limz→∞zf(z) =Aexists. Let αbe a fixed real number such that 0< α≤2π. Denote by γR the circular arc given by parametric equationz=Re^iθ, 0≤θ≤αwithR≥1.Find lim

R→∞

γ_R

f(z)dz.

Answers or Hints

13.1. (a). 1 +i¹₃, (b). ₂ⁱ

e^−iπ2/16−e^−iπ2

, (c). 0, (d). !2 1 1

2ln(1 + t²)dt+i!2

1 tan⁻¹tdt.

13.2. 8a.

13.3. !

γ|z|²dz=!2π

0 |t+it²|²(1 + 2ti)dt=!2π

0 (t²+t⁴)(1 + 2ti)dt.

13.4. !

γ(y −x−3ix²)dz = !1

0(t−t−3it²)(1 +i)dt = 1−i (γ(t) = (1 +i)t, t∈[0,1]).

13.5. !

γ z−2

z dz=!π 0 2e^iθ−2

2e^iθ 2ie^iθdθ=−4−2iπ,!

−γ|z|^1/2exp (iArgz)dz=

−!π

0 |2e^iθ|^1/2e^iθ2ie^iθdθ=−√ 2e^2iθπ

0 = 0.

13.6. !2π

0 e^i(m−n)θdθ= 2πifm=nand = "

e^i(m−n)θ/i(m−n)#^2π₀ = 0 if m=n.Now!

γz^mzⁿdz=!2π

0 e^imte^−intie^itdt= 2πiifm−n+ 1 = 0 and 0 ifm−n+ 1= 0.

13.7. In parametric form, the equation of ellipse is x = acost, y = bsint, t ∈ [0,2π]. Thus, z(t) = acost+ibsint. Since a²−b² = 1, we find√

1−z²=±(asint−ibcost) andz(t) =−asint+ibcost.

13.8. !d

cf(z₂(t))z₂(t)dt=!d

cf(z₁(φ(t)))z₁(φ(t))φ(t)dt=!b

af(z₁(s))z₁(s)ds.

13.9. !

γ dz

z2−1≤M L(γ), |z|= 2, |z²−1| ≥ |z|²−1≥3,so z2¹−1≤ ¹₃ ifz∈γ, L(γ) =^4π₄ =π.

13.10.|z²−1| ≥ |z²| −1 = 3 for|z|= 2.Thus,!

γ dz

z²−1≤^4π₃. 13.11.!

γ(e^z−z)dz≤!

γe^zdz+!

γzdz≤0 + 4(5 + 4 + 3) = 48.

·

0

−4

3i |z|= 4

13.12.Logz z²

=LogR+iArgz R²

≤^|LogR|+π

R² = LogR+π R² ,so!

γ_R

Logz z² dz≤ 2πR

LogR+π R²

= 2π

LogR+π R

. 13.13.Similar to Problem 13.12.

13.14. Let the parametric equation of γ be z = z(t), a ≤ t ≤ b. Since γ is a closed curve, z(a) = z(b). Let f(z) = u(z) +iv(z), where u, v are real, sof(z) =u_x+iv_x=v_y−iu_y.Thus,I=!b

a(u(z(t))−iv(z(t)))(u_x+ ivx)(xt+iyt)dt. Hence, ReI = !b

a(uuxxt−uvxyt+vuxyt+vvxxt)dt =

!b

a[(uuxxt+uuyyt) + (vvyyt+vvxxt)]dt= (1/2)!b

a[_dt^d(u²+v²)]dt= 0.

13.15.Clearly,!

γ_Rf(z)dz=!α

0 f(Re^iθ)Rie^iθdθ=i!α

0 Re^iθf(Re^iθ)dθand iAα=i!α

0 Adθ.Hence,|!

γ_Rf(z)dz−iAα| ≤!α

0 |Re^iθf(Re^iθ)−A|dθ.Now since limz→∞zf(z) = A, for any > 0 there exists a δ > 0 such that

|Re^iθf(Re^iθ)−A|< /αfor allθ.Therefore, it follows that|!

γ_Rf(z)dz− iAα|< ,and hence limR→∞!

γ_Rf(z)dz=iAα.

Lecture 14