Complex Analysis

Exercise 2.3. Let \(f(z) = \dfrac {\cos z}{(z^2 + 1)^2}\).

(a)

Find the singularities of \(f\).

(b)

Show that the integral of \(f\) around \(\abs {z} = 3\) is the sum of its integrals around small circles centred at the singularities.

(c)

Evaluate \(\oint _{\abs {z} = 3} f(z) \dd z\).

Solution. Singularities. \(\cos z = \frac 12(e^{iz} + e^{-iz})\) is entire and \((z^2 + 1)^2\) is a polynomial, so \(f\) is analytic except where \((z^2 + 1)^2 = 0\), that is, at \(z = \pm i\).

Decomposition. Let \(C_1\) and \(C_2\) be the positively oriented circles \(\abs {z - i} = \frac 12\) and \(\abs {z + i} = \frac 12\). They lie inside \(\abs {z} = 3\), their interiors are disjoint, and \(f\) is analytic everywhere else. By Theorem 1.3,

Around \(i\). Write \(f(z) = g_1(z) / (z - i)^2\) with \(g_1(z) = \cos z / (z + i)^2\), which is analytic on and inside \(C_1\). By Theorem 2.2 with \(n = 1\), the integral is \(2\pi i \, g_1'(i)\). Now

and \(\sin i = i \sinh 1\), \(\cos i = \cosh 1\), so

Hence \(\oint _{C_1} f(z) \dd z = 2\pi i \cdot \frac {1}{4ie} = \frac {\pi }{2e}\).

Around \(-i\). Likewise, with \(g_2(z) = \cos z / (z - i)^2\),

so \(\oint _{C_2} f(z) \dd z = -\frac {\pi }{2e}\).

Total.

Remark 2.4. The answer \(0\) can be predicted without computing. \(f\) is even, and the substitution \(z = -w\) maps the circle \(\abs {z} = 3\) onto itself with the same orientation, so \(\oint f(z) \dd z = \oint f(-w)(-\dd w) = -\oint f(w) \dd w\).

Exercise 2.5. Let \(\Gamma \) be a simple closed positively oriented contour, and let \(f\) be analytic inside and on \(\Gamma \). For \(z_0\) not on \(\Gamma \):

(a)

evaluate \(\oint _\Gamma \frac {f'(z)}{z - z_0} \dd z\) by Cauchy’s integral formula;

(b)

evaluate \(\oint _\Gamma \frac {f(z)}{(z - z_0)^2} \dd z\) by the formula for derivatives;

(c)

compare the two;

(d)

say what goes wrong when \(z_0\) lies on \(\Gamma \).

Solution. By Theorem 2.2, \(f'\) is analytic wherever \(f\) is.

\(z_0\) inside \(\Gamma \). Theorem 2.1 applied to \(f'\), and Theorem 2.2 with \(n = 1\) applied to \(f\), give

\(z_0\) outside \(\Gamma \). Both integrands are analytic on a simply connected domain containing \(\Gamma \) and its inside but not \(z_0\), so both integrals are \(0\) by Theorem 1.2.

Comparison. In either case the two integrals are equal.

\(z_0\) on \(\Gamma \). Both integrands are unbounded near \(z_0\), so they are not continuous on \(\Gamma \). Neither formula applies, and the integrals are not even defined as ordinary contour integrals; they would have to be treated as improper integrals. \(\relax \vartriangleleft \)

Remark 2.6. The equality has a one-line explanation. Off \(z_0\),

and the integral of a derivative around a closed contour is \(0\).

Exercise 3.3. Let \(f\) be entire with \(A \leq \operatorname {Im} f(z) \leq B\) for all \(z\). Show that \(e^{if}\) is bounded and entire, and deduce that \(f\) is constant.

Exercise 3.4. Let \(f\) be entire with \(f^{(7)}\) bounded on \(\C \). What kind of function is \(f\)? Generalise, and decide whether the same holds for real functions.

Solution. By Theorem 2.2, \(f^{(7)}\) is entire, and it is bounded, so it is a constant \(c\) by Theorem 3.2. Hence \(f^{(n)} \equiv 0\) for \(n \geq 8\). By Taylor’s theorem (Theorem 8.2) the series of \(f\) about \(0\) converges to \(f\) on all of \(\C \), and its coefficients \(f^{(n)}(0) / n!\) vanish for \(n \geq 8\). So

a polynomial of degree at most \(7\), and exactly \(7\) when \(c \neq 0\).

Generalisation. The same argument shows that if \(f\) is entire and \(f^{(k)}\) is bounded, then \(f\) is a polynomial of degree at most \(k\).

Real functions. This fails on \(\R \). For \(g(x) = \sin x\), every derivative is \(\pm \sin x\) or \(\pm \cos x\), so \(g^{(7)}(x) = -\cos x\) is bounded, yet \(\sin x\) is not a polynomial. \(\relax \vartriangleleft \)

Remark 3.5. Applying Theorem 3.1 directly to \(g = f^{(7)}\) gives \(\abs {f^{(8)}(z_0)} \leq M / R\) for every \(R\), which is the same argument that proves Liouville’s theorem.

Exercise 4.2. Let \(f(z) = \left (z - \frac 12\right )\left (z - \frac 14\right )\) on the closed unit disc. Explain why \(\max _{\abs {z} \leq 1} \abs {f(z)}\) is attained on \(\abs {z} = 1\), compute \(\abs {f}\) there, and find the maximum.

Solution. Where. \(f\) is a polynomial, so it is analytic on the open unit disc and continuous on its closure. By Theorem 4.1, the maximum of \(\abs {f}\) is attained on \(\abs {z} = 1\).

On the circle. Write \(z = e^{i\theta }\) and \(a = \cos \theta \). For real \(c\), \(\abs {z - c}^2 = (z - c)(\bar z - c) = 1 - 2ca + c^2\), so

Maximum. The derivative in \(a\) is \(a - \frac {27}{16} < 0\) on \([-1, 1]\), so \(\abs {f}^2\) is largest at \(a = -1\), that is, at \(z = -1\), where it equals \(\frac {225}{64}\). Hence

Remark 4.3. The triangle inequality gets there faster: for \(\abs {z} \leq 1\), \(\abs {f(z)} \leq \left (1 + \frac 12\right )\left (1 + \frac 14\right ) = \frac {15}{8}\), with equality at \(z = -1\).

Exercise 6.2. Find the radius of convergence and a closed form for \(f(z) = \sum _{n=1}^{\infty } n^3 z^n\). Use it to evaluate \(\sum _{n=1}^{\infty } n^3 / 3^n\) and \(\sum _{n=3}^{\infty } n^3 / 2^n\).

Solution. Radius. With \(a_n = n^3\),

so \(R = 1\) by Theorem 5.2.

Closed form. Start from the geometric series

Differentiating term by term (Theorem 6.1) and multiplying by \(z\) replaces each coefficient \(a_n\) by \(n a_n\) without changing the disc. Doing this three times gives

For instance, the last line comes from the quotient rule:

The sums. Both points lie inside \(\abs {z} < 1\). At \(z = 1/3\),

At \(z = 1/2\) the full sum is \(\frac {\frac 12 \cdot \frac {13}{4}}{\left (\frac 12\right )^4} = 26\), and removing the terms \(n = 1, 2\), which are \(\frac 12\) and \(2\), leaves

Remark 6.3. The operator \(\vartheta = z \frac {\dd }{\dd z}\) sends \(z^n\) to \(n z^n\), so \(f = \vartheta ^3 \bigl (\frac {1}{1-z}\bigr )\). The same trick sums \(\sum p(n) z^n\) for any polynomial \(p\).

Exercise 6.4. Let

Show that \(f\) is entire and that \(f''(z) + \frac {1}{z} f'(z) + f(z) = 0\) for \(z \neq 0\).

Remark 6.5. This \(f\) is the Bessel function \(J_0\). Multiplying through by \(z\) gives Bessel’s equation of order zero, \(z f'' + f' + z f = 0\), which then holds at \(z = 0\) as well.

Exercise 7.2. Let \(f(z) = \sum _{n=0}^{\infty } z^{2^n} = z + z^2 + z^4 + z^8 + \dotsb \). Show that \(f(z) = z + f(z^2)\), and find every power series \(g\) that satisfies \(g(z) = z + g(z^2)\) near the origin.

Solution. \(f\) satisfies the equation. The coefficients are \(0\) or \(1\), with infinitely many \(1\)s, so the radius of convergence is \(1\). For \(\abs {z} < 1\),

All solutions. Suppose \(g(z) = \sum a_n z^n\) converges near \(0\) and \(g(z) = z + g(z^2)\). Then

and by Theorem 7.1

Write \(n \geq 1\) as \(n = 2^m q\) with \(q\) odd. Halving \(m\) times gives \(a_n = a_q\), which is \(1\) if \(q = 1\) and \(0\) otherwise. So \(a_n = 1\) exactly when \(n\) is a power of \(2\), and

Conversely, each such \(g\) works, because the constant appears on both sides of \(g(z) = z + g(z^2)\). The solutions are the translates of \(f\) by constants. \(\relax \vartriangleleft \)

Remark 7.3. The gaps in the exponents \(2^n\) make \(f\) impossible to continue past the unit circle: every point of \(\abs {z} = 1\) is a singularity. This is a natural boundary, proved as a separate graduate proof.

Exercise 8.3. Find the first four nonzero terms of the Maclaurin series of \(1/\cos z\). Justify each step.

Solution. A series exists. \(\cos z\) is entire and \(\cos 0 = 1 \neq 0\), so \(1/\cos z\) is analytic near \(0\). By Taylor’s theorem it has a Maclaurin series \(\sum b_n z^n\).

Multiply out. Write \(\cos z = 1 - \frac {z^2}{2} + \frac {z^4}{24} - \frac {z^6}{720} + \dotsb \). Since \(\cos z \cdot \frac {1}{\cos z} = 1\), Theorem 8.1 gives

Compare coefficients. By Theorem 7.1 the coefficients of each power of \(z\) agree on both sides:

The odd equations read \(b_1 = 0\), \(b_3 = \frac 12 b_1 = 0\), and so on, as they must, since \(1/\cos z\) is even. Hence

Remark 8.4. The zeros of \(\cos z\) nearest the origin are \(\pm \pi /2\), so this series converges exactly on \(\abs {z} < \pi /2\). The numerators \(1, 1, 5, 61\) are the Euler (secant) numbers.

Exercise 8.5. Expand \(p(z) = z^4\) in powers of \(z - 1\).

Solution. \(p\) is entire, so by Theorem 8.2 its Taylor series about \(1\) converges to it everywhere. The derivatives at \(1\) are \(p(1) = 1\), \(p'(1) = 4\), \(p''(1) = 12\), \(p'''(1) = 24\), \(p^{(4)}(1) = 24\), and \(p^{(j)} \equiv 0\) for \(j \geq 5\). Dividing by \(j!\),

Remark 8.6. The coefficients are the binomial coefficients \(\binom {4}{j}\), as they must be: \(z^4 = \bigl (1 + (z - 1)\bigr )^4\).

Exercise 9.3. Find the Laurent series of \(f(z) = e^{1/(2z^2)}\) about \(0\), classify the singularity, and find \(\Res (f; 0)\).

Solution. The series \(e^w = \sum _{j=0}^{\infty } w^j / j!\) converges for every \(w\), because \(e^w\) is entire. Putting \(w = 1/(2z^2)\) gives, for every \(z \neq 0\),

By uniqueness of Laurent coefficients (Theorem 9.1), this is the Laurent series on \(0 < \abs {z} < \infty \). It has infinitely many negative powers, so \(0\) is an essential singularity. Only even powers occur, so \(a_{-1} = 0\) and \(\Res (f; 0) = 0\). \(\relax \vartriangleleft \)

Remark 9.4. A zero residue does not mean a mild singularity: here it sits beside an essential singularity. It only says that \(\oint f = 0\) around small circles.

Exercise 10.2. Show that \(f(z) = \dfrac {1 - \cos z}{z^2}\) has a removable singularity at \(0\), and find the value of \(f(0)\) that makes \(f\) analytic there.

Solution. For every \(z\), \(\cos z = 1 - \frac {z^2}{2!} + \frac {z^4}{4!} - \frac {z^6}{6!} + \dotsb \), so for \(z \neq 0\)

The Laurent series has no negative powers, so the singularity is removable. Equivalently, \(\lim _{z \to 0} f(z) = \frac 12\) (Theorem 10.1). Setting \(f(0) = \frac 12\) makes \(f\) the sum of an everywhere convergent power series, so \(f\) is entire. \(\relax \vartriangleleft \)

Exercise 11.2. Find the poles of \(f(z) = \dfrac {e^z}{z^2 (z - 2)^2}\) and their orders.

Solution. \(e^z\) is entire, so \(f\) is analytic except at the zeros \(0\) and \(2\) of the denominator.

At \(0\). \(f(z) = g(z) / z^2\) with \(g(z) = e^z / (z - 2)^2\), which is analytic at \(0\) with \(g(0) = \frac 14 \neq 0\). By Theorem 11.1, \(0\) is a pole of order \(2\).

At \(2\). \(f(z) = g(z) / (z - 2)^2\) with \(g(z) = e^z / z^2\), analytic at \(2\) with \(g(2) = e^2 / 4 \neq 0\). So \(2\) is also a pole of order \(2\). \(\relax \vartriangleleft \)

Exercise 12.3. Let \(f(z) = \dfrac {1}{z^2 \sin z}\). Find \(\Res (f; 0)\) in two ways, first by Theorem 12.2 and then from the Laurent series, and use it to evaluate \(\oint _{\abs {z} = 1} f(z) \dd z\).

Solution. Order of the pole. From the series for \(\sin z\),

with \(h(0) = 1 \neq 0\). So \(z^2 \sin z\) has a zero of order \(3\) at \(0\), and \(f\) has a pole of order \(3\) there (Theorem 11.1).

A series for \(z / \sin z\). This is \(1/h(z)\), which is analytic near \(0\). Write \(h = 1 - c\) with \(c = \frac {z^2}{6} - \frac {z^4}{120} + \dotsb \). For small \(z\), \(\abs {c} < 1\), and the geometric series gives

By the formula. With \(m = 3\),

differentiating term by term (Theorem 6.1).

From the Laurent series. Dividing by \(z^3\),

so again \(\Res (f; 0) = a_{-1} = \frac 16\).

The integral. The singularities of \(f\) are the zeros \(z = k\pi \) of \(z^2 \sin z\), and only \(z = 0\) lies inside \(\abs {z} = 1\). By Theorem 12.1,

Exercise 12.4. Find the singularities and the residues there of

Solution. (a) At \(0\). Since \(\frac {\sin z}{z} = 1 - \frac {z^2}{3!} + \dotsb \), \(f(z) \to 1\) as \(z \to 0\). The singularity is removable (Theorem 10.1), so there are no negative powers and \(\Res (f; 0) = 0\).

(a) At \(-1\). \(f(z) = \frac {\sin z / z}{z + 1}\), and \(\frac {\sin (-1)}{-1} = \sin 1 \neq 0\), so \(-1\) is a simple pole and

(b) At \(0\). \(g(z) = \frac {\cos z / (z - 1)^2}{z}\), with numerator \(1 \neq 0\) at \(z = 0\). This is a simple pole, and

(b) At \(1\). \(g(z) = \frac {\cos z / z}{(z - 1)^2}\), with numerator \(\cos 1 \neq 0\) at \(z = 1\), so this is a pole of order \(2\). By Theorem 12.2 with \(m = 2\),

Exercise 13.2. Evaluate \(\displaystyle \int _0^{2\pi } \frac {\dd \theta }{1 + \frac 14 \cos \theta }\).

Solution. To the circle. By Theorem 13.1, with \(C\) the circle \(\abs {z} = 1\),

Poles. The roots of \(z^2 + 8z + 1\) are \(z_{1,2} = -4 \pm \sqrt {15}\), both simple. Since \(\sqrt {15} \approx 3.87\), \(z_1 = -4 + \sqrt {15} \approx -0.13\) lies inside \(C\) and \(z_2 = -4 - \sqrt {15} \approx -7.87\) lies outside.

Residue. By Theorem 12.2,

Total.

Exercise 13.3. Show that for real \(a\) with \(0 < \abs {a} < 1\),

Solution. As above,

The roots \(z_{1,2} = \bigl (-1 \pm \sqrt {1 - a^2}\bigr )/a\) are real with product \(1\) and \(\abs {z_1} < \abs {z_2}\), so exactly one of them, \(z_1\), lies inside the circle. Its residue is \(1/(z_1 - z_2) = a / \bigl (2\sqrt {1 - a^2}\bigr )\), and so

With \(a = \frac 14\) this gives \(2\pi / \sqrt {15/16} = 8\pi / \sqrt {15}\), agreeing with Exercise 13.2. \(\relax \vartriangleleft \)

Remark 13.4. As \(a \to \pm 1\) the integral blows up: one pole moves onto the unit circle, where \(1 + a\cos \theta \) vanishes.

Exercise 14.1. Show that \(f(x) = x^{5/3}\) is differentiable on \(\R \), but \(f'\) is not differentiable at \(0\).

Solution. For \(x \neq 0\), \(f'(x) = \frac 53 x^{2/3}\). At \(0\),

which agrees with \(\frac 53 x^{2/3}\) at \(x = 0\). But

is unbounded as \(h \to 0\), so \(f''(0)\) does not exist. \(\relax \vartriangleleft \)

Remark 14.2. No analytic function behaves like this: once \(f'\) exists on an open set, so do \(f''\), \(f'''\), and the rest.

Exercise 15.1. Let \(g(z) = z^3\) and \(\omega = -\frac 12 + \frac {\sqrt 3}{2} i\). Show that \(g(1) = g(\omega )\), but \(g'\) does not vanish anywhere on the segment from \(1\) to \(\omega \).

Solution. \(\omega = e^{2\pi i/3}\), so \(g(\omega ) = e^{2\pi i} = 1 = g(1)\). If the mean value theorem held, some \(c\) on the segment would have \(g'(c) = 3c^2 = 0\), that is, \(c = 0\). But the segment consists of the points \(1 + t(\omega - 1)\), \(0 \leq t \leq 1\), with imaginary part \(\frac {\sqrt 3}{2} t\), which is zero only at \(t = 0\), the point \(1\). So \(0\) is not on the segment. \(\relax \vartriangleleft \)

Remark 15.2. The exponential gives another example: \(e^0 = e^{2\pi i}\), yet \(e^z\) is never zero.