MA231 Vector Analysis
Example Sheet 4: Hints and partial solutions
2010, term 1
Stefan Adams
A1 (b) If f=u(x,y ) + iv(x, y)then u=so that vy=ux= 0 and vx=−uy= 0. Hence vis a
constant. (c) uxx = (vy)x=vxy and uyy = (−vx)y=−vxy .
A2 (a) 2i/3. (b) 1.
A3 (a) (i) 2πie. (ii) π. (iii) 0. (b) 2πie
(n−1)! .
A4 (i) 0, (ii) π, (iii) −π, (iv) 0, (v) π.
A5 Integrate around the semicircle γ=γ1∪γ2where γ1(t) = Reit for t∈[0, π]and γ2(t) = t
for t∈[−R, R]. (a) |z2−2z+ 2|≥|z2| − |2z| − 2 = R2−2R−2for z∈γ1by the triangle
inequality. Use this to show |f(z)| ≤ 1/(R2−2R−2) for z∈γ1and hence, using the estimation
lemma, that Rγ1fdz→0as R→ ∞.Rγ2f(z) dz→R∞
−∞
cos(πx)
x2−2x+2 dx+iR∞
−∞
sin(πx)
x2−2x+2 dx.
z2−2z+ 2 = (z−(1 + i))(z−(i−i)) so that f(z)is holomorphic on and inside γexcept at
z= 1 + i. Hence by Cauchy’s integral formula Rγ= 2πi eπiz
z−(1−i)
z=1+i=−πe−π. Conclude
that R∞
−∞
cos(πx)
x2−2x+2 dx=−πe−πand R∞
−∞
sin(πx)
x2−2x+2 dx= 0.
B1 (a) Applying the CR equations to f(x+iy) = u(y) + iv(x)gives v0(x) = −u0(y)for all
x, y ∈R. Thus u0and v0are constant and thus fis of the form f(z) = ay +b+i(cx +d)for
some a, b, c, d ∈R. Using the condition from the CR equations again gives the claim. (b)
fis differentiable at the origin and nowhere else. gis differentiable on the circline around
the origin with radius 1. Both functions are nowhere holomorphic. (c) Only a=b=−1
guarantees that the function is complex differentiable on C.f−1,−1(z)=eiz.
B2 (a) parametrise the three sides of the triangle separately. From 1to i:Rγ1f=i, from ito
−1:Rγ2f=i, from −1to 1:Rγ3f= 0. Thus the result is 2i. This can be also derived
from example 15.6 in the lecture where it was shown that Rγzdz= 2i(area enclosed). (b)
Imitate the proof for R∂B(0,)f(z)/z = 2πif (0) from the lecture. (c) (i) 2πi(i3) = −2π+i6π
by Cauchy’s integral formula. (ii) Use Cauchy’s representation for the coefficient c2=f(2)(1)/2
in the power series of f(z)=ez2about z= 1.
B3 (a) the integrals are zero . . .(i) by the Cauchy integral formula, (ii) by Cauchy’s theorem, and
(iii) by the fundamental theorem of calculus (−(z−2)−2/2is a primitive). (b) Integrals are
zero for b < π/2respectively a < 1. For b > π/2resp. a > 1one gets RC
ez
(z−iπ/2)2dz=−2π
and RC
z3−4z2+sin z
(z−1)3dz=−iπ(2 + sin 1).
B4 Imitate the calculation from the lecture. Result: R∞
−∞ sin2(x)/x2dx=π.
C1 v(x, y) = p|xy|vanishes along the axes so has zero partial derivatives at the origin. The
Cauchy-Riemann equations do hold at the origin. However limr→0f(reiθ)
reiθ =i√|cos θsin θ
cos θ+isin θ
which varies as θvaries so that limz→0f(z)−f(0)
zdoes not exist.
C2 If f=u(x, y)+iv (x, y)then u2+v2= 0. Differentiate to find uux+vvx= 0 and uuy+vvy= 0.
Combine these with the Cauchy-Riemann equations to show that uand vare constants.
C2 f0(z)is also holomorphic on Cand so from question A4 we know that f0(z)is a linear function
and hence that fis a quadratic, namely f(z) = f(0) + zf 0(0) + z2(f00 (0)/2). The hypothesis
implies that f0(0) = 0. Also by applying Cauchy’s integral formula to f0(z)we have that
|f00(0)| ≤
1
2πi R∂B (0,1)
f0(z)
z2dz
≤1.
C3 (a) Use the quotient rule for differentiating. (b) By the uniqueness theorem (1+ z)kmust have
the same power series inside |z|<1as the real power series known via the binomial theorem
on R.
