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Problems for MATH-
Complex Analysis
Gregor Kovaˇciˇc
December 1, 2007
This list will change as the semester goes on. Please make sure you
always have the newest version of it.
- Prove the following
Theorem 1 For every power series
n=0 anz
n (^) there exists a number R, 0 ≤ R ≤ ∞, called
the radius of convergence, with the following properties:
(i) The series converges absolutely for every z with |z| < R. If 0 ≤ ρ < R the convergence is uniform for z ≤ ρ.
(ii) If |z| > R the terms of the series are unbounded, and the series is consequently divergent.
(iii) In |z| < R the sum of the series is an analytic function. The derivative can be obtained by termwise differentiation, and the derived series has the same radius of convergence.
The radius of convergence is given by the formula
1 R
= lim n→∞ sup |an| n^1 .
HINTS: (a) To prove convergence and uniform convergence, use an appropriately chosen geometric series as a majorant.
(b) To prove divergence, show that there are infinitely many unbounded terms.
(c) To prove that the derived series
n=1 nanz
n (^) has the same radius of convergence, first
prove that limn→∞ n n^1 = 1 by setting n
(^1) n = δn, and using the first two terms of the binomial theorem to show that δ n^2 < 2 /n.
(d) Write f (z) =
n=0 anz n (^) = sn(z) + Rn(z), where sn(z) = a 0 + a 1 z + · · · + an− 1 zn− (^1) , and
Rn(z) =
k=n akz
k. Write also f 1 (z) = ∑∞ n=1 nanz
n (^) = limn→∞ sn(z). Use the identity
f (z) − f (z 0 ) z − z 0
−f 1 (z 0 ) =
sn(z) − sn(z 0 ) z − z 0
− s′ n(z 0 )
+(s′ n(z 0 ) − f 1 (z 0 ))+
Rn(z) − Rn(z 0 ) z − z 0
for z 6 = z 0 and |z|, |z 0 | < ρ < R. Estimate the three terms in this identity, in particular, estimate the last term by the inequality
∣ ∣ ∣ ∣
Rn(z) − Rn(z 0 ) z − z 0
∑^ ∞
k=n
k |ak| ρk.
- We say that an open subset of the complex plane is connected if it cannot be decomposed into two disjoint open sets. Prove
Theorem 2 A nonempty open set in the plane is connected if and only if any two of its points can be joined by a polygon whose sides are parallel to the coordinate axes.
- Show the Schwartz lemma:
Theorem 3 If f (z) is analytic for |z| < 1 and satisfies the conditions |f (z)| ≤ 1 , f (0) = 0, then |f (z)| ≤ |z| and |f ′(0)| ≤ 1. Equality holds if and only if f (z) = cz with a constant c of absolute value 1.
HINT: Apply the maximum principle to the function f 1 (z) which is equal to f (z)/z for z 6 = 0 and f ′(0) for z = 0.
- Let f (z) be analytic in the closed domain bounded by the contour C; z 1 , z 2 ,.. ., zn are arbitrary distinct points within C and ωn(z) = (z − z 1 )(z − z 2 )... (z − zn). Show that the integral
P (z) =
2 πi
C
f (ζ) ωn(ζ)
ωn(ζ) − ωn(z) ζ − z
dζ
is a polynomial of degree (n − 1) which is equal to f (z) at the points z 1 , z 2 ,.. ., zn.
- The Hermite polynomials Hn(z) are defined by the expansion
e^2 tz−t
2
∑^ ∞
n=
Hn(z) n!
tn.
Prove the following relations:
(d) Show that, when n is an integer,
Jn(y + z) =
∑^ ∞
n=−∞
Jm(y)Jn−m(z).
(e) Show that if r^2 = x^2 + y^2
J 0 (r) = J 0 (x)J 0 (y) +
∑^ ∞
n=
(−1)nJ 2 n(x)J 2 n(y).
- (a) Given the function
A(z) =
z
e−^1 /t t^2
dt
find a Laurent expansion in powers of z for |z| > R, R > 0. Why will the same procedure fail if we consider
E(z) =
z
e−t t
dt?
(b) A formal series for E(z) in part (a) can be obtained by repeated differentiation by parts, that is,
E(z) =
e−z z
z
e−t t^2
dt
e−z z
e−z z^2
z
2 e−t t^3
dt = · · ·
If this procedure is continued, show that the result is
E(z) =
e−z z
z
(−1)nn! zn
with
Rn(z) = (−1)n+1(n + 1)!
z
2 e−t tn+^
dt.
Explain why the series obtained as n → ∞ does not converge.
(c) In (b), consider z = x real. Show that
|Rn(x)| ≤ (n + 1)!
e−x x^2
Explain how to approximate the integral E(x) for large x, given some n, and why this approximation holds true for Re z > 0. Why does the approximation fail as n → ∞?
- Investigate the singularities of the function
f (z) = log
z + 1 z − 1
- Let f (z) be analytic and single-valued in the ring-shaped region r 1 ≤ |z| ≤ r 3. Let |z| = r 2 be a circle satisfying r 1 < r 2 < r 3. Denote the maximun of f (z) on rj by M(rj ), j = 1, 2, 3. Then prove Hadamard’s three-circles theorem, which states that log M(r) is a convex function of log r, in the sense that
log M(r 2 ) ≤
log r 3 − log r 2 log r 3 − log r 1
log M(r 1 ) +
log r 2 − log r 1 log r 3 − log r 1
log M(r 3 ).
When would equality occur?
HINT: Consider w(z) = zsf (z), where s is real. Since |z|s^ is single-valued, the maximum- modulus theorem is easily seen to apply. Then choose s appropriately.
- Draw the Riemann surface of the functions
f (z) =
[
(x − 1)(x − 2)(x − 3)(x − 4)(x − 5)
] 12
and
g(z) =
[
(x − 1)(x − 2)(x − 3)(x − 4)(x − 5)(x − 6)
] 12
Topologically, is there any difference between the two?
- Show that the power series
∑^ ∞
n=
(−1)n−^1
(z − 1)n n
and iπ −
∑^ ∞
n=
(z + 1)n n
are analytic continuations of one another.
HINT: Both lie on the same branch of log z, so just find an appropriate path.
- Show that the function
f (z) =
∑^ ∞
n=
zn!
tends to infinity along every radius whose argument is a rational multiple of π, and thus has the unit circle as a natural boundary.
HINT: Show that
L(w, m) =
wm w − 1
and so for |w| < 1 /2 the estimate L(w, m) ≤ 2 |w|m^ holds. If |z| < R and for all j such that |zj | < 2 R, deduce that aj zj
L
z zj
, m
2 Rm|aj | |zj |m+^
- Euler’s gamma function Γ(z) is defined by the integral
Γ(z) =
0
e−ttz−^1 dt.
(a) Show that Γ(z) is analytic at least in the half-plane z > 0, that Γ(z + 1) = zΓ(z), and that Γ(n + 1) = n!.
(b) Using
e−t^ = lim n→∞
t n
)n
prove
Γ(z) = lim n→∞ nz
0
(1 − τ )nτ z−^1 dτ.
Justify the interchanging of limits. Then integrate by parts to obtain
Γ(z) =
z
∏^ ∞
1
[(
n
)z (^) ( 1 +
z n
)− 1 ]
(c) Show that
γ = lim n→∞
n
− log n
exists.
(d) Use (b) and (c) to show that
Γ(z)
= zeγz
∏^ ∞
1
[(
z n
e−z/n
]
Conclude that Γ(z) is meromorphic in the entire plane, with poles at z = −n, n = 0, 1 , 2 , 3 ,.. ., and is without zeros.
(e) Reason that the formula Γ(z + 1) = zΓ(z) holds for all z except z = −n, n = 0, 1 , 2 , 3 ,.. ..
(f) Use (a), (d), and the infinite product expansion of sin z to show that
Γ(z)Γ(1 − z) =
π sin πz
Deduce that
Γ
π.
(g) Show that the residue of Γ(z) at z = −n is (−1)n/n!.
(h) Show that
d dz
Γ′(z) Γ(z)
∑^ ∞
n=
(z + n)^2
(i) Show that Γ(z)Γ(z + 12 ) and Γ(2z) have the same poles and use (a), (f), and (h) to deduce that
Γ(z)Γ
z +
= 2^2 z−^1 Γ(2z).
(j) Show that
Γ(z) =
2 i sin πz
C
tz−^1 et^ dt, (1)
where C is the path in the figure encircling the origin, for all z not equal to a positive integer.
HINT: First assume Re z > 0. For large negative real parts of z, estimate the integral along short vertical segments emanating from the negative real axis. Conclude that the integration path C can be deformed into the union of the two half-rays {ve±iπ^ | ρ < v < ∞}, taken in the appropriate directions, and the circle {ρeiθ^ | −π < θ < π}. Show that the integrals on the two rays give you (1), and that the integral along the circle vanishes as ρ → 0. Then argue that the integral on the right-hand side of (1) must exist for all z.
(k) Use (j) and (f) to show that
1 Γ(z)
i 2 π
C
tz−^1 et^ dt.
(l) Compute (^) ∫ ∞
−∞
e−x 2 dx.
HINT: Use
f (z) = cot z −
z
and let Γn be the square with corners (n + 1/2)π(± 1 ± i).
(c) Deduce from (b) that
sin z = z
n 6 =
z nπ
ez/nπ^ = z
∏^ ∞
n=
z^2 n^2 π^2
(d) Express explicitly the function represented by the product
∏^ ∞
n=
z n^2 π^2
- The Riemann zeta function is defined as
ζ(s) =
∑^ ∞
n=
ns^
, s = σ + iτ.
(a) If Re s > 1, show that
1 ζ(s)
∏^ ∞
n=
psn
where {pn} is the ascending sequence of all the prime numbers.
HINT: To show the convergence of the product on the right-hand side, use the appropriate theorem linking this convergence to the convergence of an appropriate series. To show the equality, show that
ζ(s)(1 − 2 −s)(1 − 3 −s) · · · (1 − p−Ns ) =
m−s,
where the sum runs over all the integers that contain none of the prime factors 2, 3,... , pN.
(b) In (a), we have tacitly assumed that there are infinitely many primes. If you do not assume this fact, reason as in (a) to prove it.
HINT: If not, show that you would have lims→ 1 ζ(s) < ∞.
(c) Show that
Γ(s)ζ(s) =
0
xs−^1 ex^ − 1
dx.
HINT: Replace x by nx in the integral
Γ(s) =
0
xs−^1 e−x^ dx
and sum the appropriate series. Justify the
(d) Show that
ζ(s) = −
Γ(1 − s) 2 πi
C
(−z)s−^1 ez^ − 1
dz,
where (−z)s−^1 is defined on the complement of the real axis as e(s−1) log(−z)^ with −π < Im m log(−z) < π, and C is the path shown in the figure.
HINT: Let the radius of the circle go to 0. Also, somewhere in the calculation, use part (c) and the formula Γ(s)Γ(1 − s) = π/ sin πs.
2 π i
(e) Conclude that the ζ-function is meromorphic in the whole plane and its only pole is a simple pole at s = 1 with residue 1.
HINT: Argue that the zeros and poles of Γ(s) and the integral on the right-hand side of the formula in part (d) cancel when s is a positive integer. At s = 1, use problem 20 (g) and compute the integral by residues to show that it is 1.
- Determine explicitly the largest disks about the origin whose respective images under the mappings w = z^2 + z and w = ez^ are one-to-one.
- Use the open mapping theorem to prove the maximum modulus theorem.
- Use Rouch´e’s theorem to show the fundamental theorem of algebra.
- (i) Let w = f (z) be analytic near the point z = z 0 and let f (z 0 ) = w 0. If f ′(z 0 ) 6 = 0, show that the inverse function z = f −^1 (w) is given by
z = f −^1 (w) = z 0 +
∑^ ∞
n=
n!
dn−^1 dzn−^1
[
ψ(z)
]n}
z=z 0
(w − w 0 )n,
with λ ∈ R and |a| < 1.
HINT: Inverse points a and 1/¯a must be mapped into 0 and ∞, respectively.
(ii) Show that the most general linear fractional transformation of the circle {|z| < ρ} onto the circle {|w| < r} is given by
w = ρr eiλ^
z − a ¯az − ρ^2
with λ ∈ R and |a| < ρ.
(iii) Show that the most general linear fractional transformation that maps the upper half- plane Im z > 0 onto the unit circle {|w| < 1 } is given by the formula
f (z) = eiλ^
z − a z − a¯
where λ ∈ R and Im a > 0.
HINT: Conjugate values of some z must map into 0 and ∞, respectively.
- (i) Show that a one-to-one conformal mapping of {|z| < 1 } onto itself that preserves the origin must be a linear function of the form f (z) = eiαz, with α ∈ R.
HINT: Use the Schwartz lemma (Problem 3) and the fact that an analytic function with constant modulus must be a constant.
(ii) Use (i) to show that any one-to-one conformal mapping of {|z| < 1 } onto itself must be a linear fractional transformation.
- Prove the following
Theorem 5 If f (z) is analytic for |z| < 1 , Re[f (z)] > 0 , and f (0) = a > 0 , then f ′(0) ≤ 2 a.
HINT: Compose f with a linear fractional transformation φ so that the function g = φ ◦ f maps {|z| < 1 } onto itself and g(0) = 0. Cauchy’s formula on circles with radii 1 − ε shows |g′(0)| ≤ 1. Invert φ to find f (z) in terms of g(z), and relate the corresponding derivatives.
- Show that the mapping
w = −
R
α
z − α z − β
for some real α and β, maps the region contained between the circles |z − a| = r and |z| = R, 0 < a < R − r, conformally onto some annulus ρ < |w| < 1. Compute that
α =
2 a
R^2 + a^2 − r^2 − A
, β =
R^2
α
where A =
R^2 + a^2 − r^2
− 4 a^2 R^2 ,
and
ρ =
R
α
α − a β − a
HINT: The equation |z − α| = k|z − β|, k > 0 is the equation of a circle with respect to which α and β are symmetric to one-another. You can choose α, β ∈ R (why?) and vary k so that |z − a| = r and |z| = R are part of a family of circles with this property. This will let you compute α and β. Finally, choose k = −R.
- Prove the following
Theorem 6 The only one-to-one conformal mapping of the extended plane onto itself is the linear fractional transformation.
HINT: First show that a one-to-one conformal mapping f (z) of the finite plane onto itself must be the linear transformation as follows:
(i) Argue that f ′(z) 6 = 0 in C, since f (z) − f (z 0 ) would have at least two distinct roots near any z 0 with f ′(z 0 ) = 0.
(ii) Use the open mapping theorem to show that the image of |z| < 1 contains some disk |w − w 0 | < A and deduce from the Weierstrass theorem that ∞ cannot be an essential singularity for f (z).
(iii) The function f (z) must have a pole at ∞, and so must be a polynomial. Since f ′(z) 6 = 0, it must be linear.
(iv) Show that the general case follows by applying the transformation ζ = 1/(z − z 0 ), where z 0 is the points that is mapped to ∞.
- Prove that in any region Ω the family of analytic functions with positive real part is normal. Under what added condition is it locally bounded?
HINT: Consider the functions e−f^.
- Show that the functions zn, where n is a nonnegative integer, form a normal family in |z| < 1 and also in |z| > 1, but not in any region that contains a point on the unit circle.
- If the domain Ω contains the point at infinity and is mapped onto the interior of the unit circle in such a way as to make the points z = ∞ and w = 0 correspond to each other, show that the mapping function w = f (z) has the expansion
f (z) =
∑^ ∞
n=
an zn^
, a 1 6 = 0,
which converges in the exterior of a sufficiently large circle.
- A fluid flows with initial velocity u 0 through a semiinfinite channel of width d and emerges through the opening AB of the channel (see the top figure). Find the speed of the fluid at any point.
HINT: First show that the conformal mapping w = z + e^2 πz/d^ maps the channel |y| < d/ 2 onto the w-plane excluding slits, as indicated in the bottom figure.
- Derive the complex potential Ω(z) for flow past an elliptic cylinder with semiaxes a and b, where the flow has velocity U + iV at infinity, and circulation γ.
HINT: First show that
w =
a + b
z +
z^2 − c^2
, c^2 = a^2 − b^2 ,
(choose the positive square root) maps the exterior of the ellipse with semiaxes a and b onto the exterior of the circle |w| = 1.
ANSWER:
Ω(z) =
γ 2 πi
log
z +
z^2 − c^2
U
a − b
az − b
z^2 − c^2
iV a − b
bz − a
z^2 − c^2
- Find the flow past an infinitely long and infinitely thin vertical wall of height s, if the velocity far away from this wall is u 0 in the horizontal direction.
HINT: Use the Schwartz-Christoffel transformation to show that the upper half-plane with a “slit” between 0 and is is mapped onto the upper half of the z plane by the (inverse of the) function w = s
z^2 − 1.
- Consider Bessel’s equation
z^2 w′′^ + zw′^ + (z^2 − ν^2 )w = 0.
(i) Show that one solution is always the Bessel function of order ν,
Jν (z) =
∑^ ∞
n=
(−1)n n!Γ(ν + n + 1)
( (^) z 2
) 2 n+ν .
Show that, if ν is not an integer of a half-integer, a second linearly independent solution is J−ν (z).
(ii) For the Bessel function Jν (z), show the relations
Jν− 1 (z) + Jν+1(z) =
2 ν z
Jν (z),
and d dz
[
z−ν^ Jν (z)
]
= −z−ν^ Jν+1(z).
Conclude by induction that
z−ν−nJν+n(z) = (−1)n
d zdz
)n [ z−ν^ Jν (z)
]
(iii) If ν is a non-negative integer, show that a second linearly independent solution is the Neumann function of order ν,
Yν (z) = 2Jν (z)
[
log
(z 2
]
∑^ ν−^1
n=
(ν − n − 1)! n!
( (^) z 2
)−ν+2n
∑^ ∞
n=
(−1)n n!(ν + n)!
(ν+n ∑
k=
k
∑^ n
k=
k
(z 2
) 2 n+ν .
Show also that
Yν (z) = lim μ→ν
Jμ(z) cos πμ − J−μ(z) sin πμ
(iv) Show that
J^12 (z) =
πz
sin z, J− 12 (z) =
πz
cos z,
