MA231 Vector Analysis
Example Sheet 3: Hints and partial solutions
2010, term 1
Stefan Adams
A1 ∇ · v= 4rsin φ−1
r2tan φand ∇ × v=rcos φˆeθ.
A2 (a) i:r= 1,θ =π/2;−i:r= 1, θ =−π/2;1 + i:r=√2, θ =π/4;−1
2+√3
2i:r= 1, θ =
2π/3. (b) 3
20 −1
20 i;−i;−1
2+√3
2i. (c) 1+3iand −1−3i.1+ (3−√8)iand −1−(3+ √8)i.
(d) cos θ+isin θfor θ=π/4,3π/4,5π/4,7π/4.
A3 (a) ¯z=x−iy,|¯z|=|z|and arg(¯z) = −arg(z). (b) split into real and imaginary parts.
A4 Since |z|=z¯zthe set is the unit circle and we can use φ(t)=e2πit.
A5 (a) (i) R=∞, (ii) R=e(recall (1 + 1
n)n→e). (b) The series converges on the set
{z:|z−4i|<1/5}and nowhere else. (c) |in
i+n2|=|1
i+n2| ≤ 1
n2and the first series converges
absolutely. The real part of 1
i+nis n
n2+1 so the real part of the second series diverges by
comparison with P1
2n.
A6 (a) Use the definition of exp(z)and sin(z)as power series. (b) Use part (a) or argue (after
week 9) that the identity is true for real zand since both sides are holomorphic functions
on Cthe identity must remain true on the whole of C(by Taylor’s theorem for holomor-
phic functions and the uniqueness theorem for power series). (c) Use the representations for
sin,cos,sinh,cosh in terms of the exponential function.
B1 (a) We can choose any surface Swith boundary C, simplest is r(u, v)=(ucos v, u sin v , 1−
ucos v−usin v)for u∈[0,1], v ∈[0,2π]. Surface normal: N=∂r
∂u ×∂r
∂v = (u, u, u)
(using a picture we can see that this is the correct orientation of N). ∇ × v= (0,0,3u2),
RChv, ˆ
Ti=RSh∇ × v, ˆ
Ni= 3π/2.
(b) The relations kˆerk=kˆeφk=kˆezk= 1 as well as ˆer·ˆeφ= ˆer·ˆez= ˆeφ·ˆez= 0 and
ˆer׈eφ= ˆezare easily checked by direct calculation.
(c) ∇T(r, φ, θ) = ∂T
∂r ˆerand ∆T(r, θ, φ) = 1
r2
∂
∂r r2∂T
∂r .
B2 (a) Solution: all z∈Cwith |z−(1
3+4
3i)|>√8
3. (b) z7= 8 −8i.
B3 (a) Convergence inside the circle of radius 1/6centered at −i/3and divergence outside this
circle (using the ratio test). Checking the Cauchy criterion gives convergence on the boundary.
(b) |1
z+n2| ≤ 1
|Re(z+n2)|≤1
n2when zhas positive real part. So the series converges absolutley
for all such zand, as the uniform limit of continuous functions, it is continuous. (c) (i)
z2= (z−i+i)2=−1(z−i)0+ 2i(z−i)1+ (z−i)2,R=∞. (ii) Use 1
z−1=1
i−1+z−i=
1
i−1(1 + z−i
i−1)−1and then the geometric series expansion of (1 + w)−1.R=√2. (d) Write
a quadratic equation in ezand get zas the corresponding complex logarithm.
B4 (a) Use the rewriting of sin and cos via the exponential function. (b) Use properties of M¨obius
transformations (see lecture). In particular you may use the following fact: If fmaps the circle
|z−a|=ronto the circle |z−b|=ρ, then fmaps the closed disk |z−a| ≤ rto either
|z−b| ≤ ρor |z−b| ≥ ρ.
C2 The components mi
rof usatisify ∆(mi/r)=0. Also ∇ · m
r=1
rm·x. The identity then shows
the result.
C3 (c) |fB(x)| ≤ C
|x|−RVol(B(0, R)) →0as |x|→∞. (d) A= lim
|x|→∞ −CRB(0,R)|x|
|x−y|dV(y).
Now take the limit inside the integral.
C4 (a) Recall that z w = ¯z¯wso that zn=zn. Since the coefficients are all real we have 0 =
p(z) = p(z). (c) Multiply out (z−zj)(z−¯zj)when zj=aj+ibj.