Cauchy’s Integral Formula for Derivatives
Theorem 18.6 Liouville’s Theorem). The only bounded entire functions are the constant functions
Proof.
Supposef is analytic and bounded by some numberM over the whole complex plane C. By Theorem 18.5, for the case n = 1, we have|f(z0)| ≤M/R for any z0 ∈ C and for any R > 0.Letting R → ∞, we findf(z0) = 0 for anyz0.Hence,f vanishes everywhere; i.e.,f must be a constant.
Problems
18.1. Show that the Cauchy integral formula implies the Cauchy- Goursat Theorem.
18.2. Suppose thatf is analytic in a simply connected domainS,and letz0∈Sbe a fixed point. Define the integral along any contour connecting the pointsz0andz∈SasF(z) =
z z0
f(ξ)dξ.Show thatF(z) =f(z); i.e., the integral is also an analytic function of its upper limit.
18.3. Suppose thatf is analytic in a domain S, and let z0 ∈ S be a fixed point. Define
g(z) =
⎧⎨
⎩
f(z)−f(z0)
z−z0 if z=z0 f(z0) if z=z0. Show thatg is analytic inS.Hence, deduce that the function
g(z) =
⎧⎨
⎩ sinz
z if z= 0 1 if z= 0
(18.7)
Cauchy’s Integral Formula for Derivatives 121 is entire.
18.4. Let f be analytic within and on a positively oriented closed contourγ,and the pointz0 is not onγ.Show that
γ
f(z)
(z−z0)2dz =
γ
f(z) z−z0dz.
18.5. Letu=u(x, y) be a harmonic function in a domainS.Show that all partial derivativesux, uy, uxx, uxy, uyy,· · ·exist and are harmonic.
18.6. Letγbe a simple closed contour described in the positive sense, and writeg(z) =
γ
ξ3+ 7ξ
(ξ−z)3dξ.Show that g(z) = 6πiz whenz is insideγ and thatg(z) = 0 whenzis outsideγ.
18.7. Let f(z) = (ez+e−z)/2. Evaluate
γ
f(z)
z4 dz, where γ is any simple closed curve enclosing 0.
18.8. (a). Letγbe the contourz=eiθ, −π≤θ≤πtraversed in the positive direction. Show that, for any real constanta,
γ
eaz
z dz = 2πi.
(b). Deduce that π
0
eacosθcos(asinθ)dθ=π.
18.9. Letγ denote the boundary of the rectangle whose vertices are
−2−2i,2−2i,2 +iand−2 +iin the positive direction. Evaluate each of the following integrals:
(a).
γ
e−z
z−πi4 dz, (b).
γ
cosz
z4 dz, (c).
γ
z (2z+ 1)2dz, (d).
γ
e−z
z2+ 2dz, (e).
γ
dz
z(z+ 1), (f).
γ
ezsinz+ 1 (z2+ 3)2
dz.
18.10.Letf be analytic inside and on the unit circleγ.Show that, for 0<|z|<1,
2πif(z) =
γ
f(ξ) ξ−zdξ−
γ
f(ξ) ξ−1/zdξ.
Hence, deduce thePoisson integral formula f(reiθ) = 1
2π 2π
0
(1−r2)
1−2rcos(θ−t) +r2f(eit)dt, 0< r <1.
18.11.Iff(z) is analytic and|f(z)| ≤1/(1− |z|) in |z|<1, show that
|f(0)| ≤4.
18.12. Let the function f be entire andf(z) → ∞ as z → ∞. Show that f must have at least one zero.
18.13.Letf(z) be an entire function such that|f(z)| ≤ |z|.Show that f(z) =a+bz2with some constantsa, b∈Csuch that |b| ≤1.
18.14.Supposef(z) is an entire function withf(z) =f(z+1) =f(z+i) for allz∈C.Show thatf(z) is a constant.
18.15.Supposef(z) andg(z) are entire functions,g(z)= 0 and|f(z)| ≤
|g(z)|, z ∈C.Show that there is a constantc such thatf(z) =cg(z).
18.16. Show that an entire function whose real part is nonpositive is constant.
18.17. A subsetS of C is said to be densein C if its closure S =C.
Show that the image of a nonconstant entire function is dense inC.
18.18. Suppose f(z, t) is continuous, and continuously differentiable with respect to t,and supposeγ is a smooth curve in the domain of defi- nition off.Then, the functionF(t) =
γ
f(z, t)dzis continuously differen- tiable with respect tot,and its derivative is
dF(t)
dt =
γ
∂f(z, t)
∂t dz.
18.19.Letf(z, λ) be a continuous function ofzon the bounded closed regionS for each value ofλin |λ−λ0|< ρ,andf(z, λ)→F(z) uniformly asλ→λ0.Show thatF(z) is continuous onS.Moreover, ifγis any closed contour lying inS,then
lim
λ→λ0
γ
f(z, λ)dz =
γ
F(z)dz.
18.20. Let f(z, λ) be an analytic function of z on the domain S for each value ofλin|λ−λ0|< ρ,andf(z, λ)→F(z) uniformly asλ→λ0 in every closed regionGof S.Show thatF(z) is analytic onS. Moreover, as λ→λ0, ∂f(z, λ)/∂z→F(z) uniformly onG.
Answers or Hints
18.1. Letf be a complex function that is analytic throughout the region enclosed by a simple closed contour γ. Let z0 be a point inside γ. Then
!
γf(z)dz=!
γ
f(z)(z−z0)
z−z0 dz= 2πi(f(z0)(z0−z0)) = 0.
Cauchy’s Integral Formula for Derivatives 123 18.2. Ifz+ Δz∈S,then F(z+Δz)Δz−F(z)−f(z) = Δz1 !z+Δz
z [f(ξ)−f(z)]dξ.
Since f is continuous at z, |ξ−z|< δ, or in particular |Δz| < δ implies
|f(ξ)−f(z)|< .Thus,F(z+Δz)Δz−F(z)−f(z)<|Δz |!z+Δz
z |dξ|=. 18.3. LetB(z0, R) be an open disk contained inS.Forz=z0in B(z0, R) consider the segmentξ(t) =z0(1−t) +tz, t∈[0,1].Sincef is analytic in B(z0, R) we can writeg(z) asg(z) = !1
0 f((z−z0)t+z0)dt. This is also defined atz =z0, in fact,g(z0) = !1
0 f(z0)dt=f(z0). Now use the fact thatf is analytic inB(z0, R).
18.4. Use (17.1) forf(z) and (18.6) forn= 1.
18.5. For each point z0 = (x0, y0) inS there exists a diskB(z0, R)⊂S.
In this disk, a conjugate harmonic functionv exists, so that the function f =u+iv is analytic. Now use Theorem 18.1.
18.6. If z is inside γ, we have !
γ ξ3+7ξ
(ξ−z)3dξ = 2πi2! f(z) = 6πiz (f(ξ) = ξ3+ 7ξ). If z is outside γ, then (ξξ3−+7ξz)3 is analytic throughout the region enclosed byγ(including γ) and therefore!
γ ξ3+7ξ
(ξ−z)3dξ = 0.
18.7.0.
18.8. (a). 2πi(ea·0),(b).!
γ eaz
z dz=!π
−π
eacosθeiasinθ eiθ ieiθdθ
=!π
−πieacosθcos(asinθ)dθ−!π
−πeacosθsin(asinθ)dθ= 2πi.
18.9. (a). 2πi(e−πi/4), (b). 0,(c). πi/2,(d).−πe√2i/√
2,(e). 0, (f).−π√
3/18.
18.10.1/zis outsideγ.Takez=reiθ andξ=eit.
18.11. For anyr <1, letγr denote the circle|z|=r. By (18.6), we have
|f(0)| ≤ 2π1 !
γr
(1−|z1|)|z|2|dz| ≤ 2π1 r2(11−r)2πr= r(11−r).Now letr= 1/2.
18.12. If f has no zero in C, then g(z) = 1/f(z) is entire and g(z)→0 as z→ ∞. Now show thatg is bounded, and then apply Theorem 18.6 to conclude thatgis a constant.
18.13. Note that f(z0) = 2πi1 !
γR[f(z)/(z−z0)2]dz, where γR is the circle|z−z0|=Rtaken in the positive direction. Hence,|f(z0)| ≤MR/R where MR ≥ |f(z)| on |z−z0| = R. Now, since |f(z)| ≤ |z|, it follows that |f(z0)| ≤ (R+|z0|)/R for all R. Thus, in particular, for all z0 ∈ C, |f(z0)| ≤ 2. Therefore, in view of Theorem 18.6, f(z) must be a constant, and so f(z) = a+αz+bz2. However,|f(z)| ≤ |z|implies that f(0) = 0, and hence we should haveα= 0; i.e., f(z) =a+bz2. Finally, since|f(z)|=|2b| ≤2, |b| ≤1.
18.14.Observe thatf(z) is bounded, and apply Theorem 18.6.
18.15.Observe thatf(z)/g(z) is entire and|f(z)/g(z)| ≤1.Now use The- orem 18.6.
18.16. Iff =u+iv,then u≤0 implies |ef|=eu ≤1. Thus, in view of Theorem 18.6,ef =c.Hence,eff = 0,which givesf = 0.
18.17. Letf : C →C be an entire function. Iff(C) is not dense in C, then C\f(C) is open and =∅. Thus, there exist w∈C\f(C) and r >0 such that B(w, r) ⊂ C\f(C). Hence, |f(z)−w| ≥ r > 0 for all z ∈ C.
Consider the functiong :C→Cdefined by g(z) = 1/(f(z)−w).Clearly, g(z) is entire sincef(z) is; moreover,|g(z)| ≤1/r,so g(z) is bounded. By Theorem 18.6,g(z) must be a constant, and hencef(z) must be a constant.
18.18.ConsiderF(t)t−−F(tt 0)
0 = t−1t
0
!
γ[f(z, t)−f(z, t0)]dz.Iffis real-valued, then by the Mean Value Theorem we have F(t)t−−F(tt 0)
0 =!
γ
∂f(z,t∗)
∂t dz,where t∗lies betweent0 andt but depends onz.Thus,
F(t)t−−F(tt0 0)−!
γ
∂f(z,t)
∂t dz≤!
γ
∂f(z,t∂t∗)−∂f(z,t)∂t |dz|.
Now, using uniform continuity of ∂f∂t in γ×[t0, t] and letting t →t0, the result follows. If f is complex-valued, we use the same argument for the real and imaginary parts separately. Finally, the continuity of ∂f∂t implies that of ∂F∂t.
18.19.Sincef(z, λ) is continuous, for every >0 there exists aδ=δ(z0, λ) such that |z−z0|< δ ⇒ |f(z, λ)−f(z0, λ)|< /3. Since f(z, λ)→F(z) uniformly, there exists 0 < ρ ≤ ρ such that |λ−λ0| < ρ ⇒ |f(z, λ)− F(z)|< /3.Thus,|z−z0|< δ, |λ−λ0|< ρ ⇒ |F(z)−F(z0)| ≤ |F(z)− f(z, λ)|+|f(z, λ)−f(z0, λ)|+|f(z0, λ)−F(z0)|< .Again, sincef(z, λ)→ F(z) uniformly, for every >0 there exists 0< ρ≤ρsuch that|λ−λ0|<
ρ⇒ |f(z, λ)−F(z)|< /L(γ),but then !
γ(f(z, λ)−F(z))dz< . 18.20.Similar to that of Problem 18.19.