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Mathematical Methods

Solved exercises on the complex integral and Cauchy's integral formula

Nine exercises with solutions and contours drawn: holomorphic and non-holomorphic functions, integral formula, derivative formula, ML estimates, Cauchy's estimate, Liouville, maximum modulus.

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Exercises on the complex integral, Cauchy’s theorem and Cauchy’s integral formula of the kind set in Mathematical Methods and Complex Analysis courses, in three levels: basic, with direct computation and Cauchy’s theorem; intermediate, with the integral formula and the formula for derivatives; exam, with estimates, Liouville’s theorem and the maximum modulus principle. The result is given straight away; the full solution, with the contour drawn, opens with a click. The proofs are in the in-depth study linked at the top; real integrals computed with residues are on a separate exercise page.

Basic level

Exercise 1. Holomorphic or not

Compute \oint_{|z| = 1}\bar z\,dz and \oint_{|z| = 1}z^2\,dz, traversing the circle anticlockwise. Why are the results different?

Result: 2\pi i and 0.

Solution

With z = e^{it}, dz = ie^{it}dt, \bar z = e^{-it}:

    \[ \oint\bar z\,dz = \int_0^{2\pi}e^{-it}\,ie^{it}dt = 2\pi i,\qquad \oint z^2\,dz = \int_0^{2\pi}e^{2it}\,ie^{it}dt = i\int_0^{2\pi}e^{3it}dt = 0 \]

z^2 is holomorphic (it has the antiderivative z^3/3), and Cauchy’s theorem guarantees a zero integral. \bar z is not holomorphic: with z = x + iy, \bar z = x - iy has u = x, v = -y, and the Cauchy-Riemann condition u_x = v_y gives 1 = -1, false. On the unit circle \bar z = 1/z, which is why the integral coincides with that of 1/z.

The unit circle
The integration contour, traversed anticlockwise.

Exercise 2. The integral formula

Compute \displaystyle\oint_{|z| = 2}\frac{\sin z}{z - \pi/2}\,dz.

Result: 2\pi i.

Solution

\sin z is holomorphic everywhere and the point \frac{\pi}{2} \approx 1.57 is inside the circle of radius 2. Integral formula with f = \sin z, z_0 = \frac{\pi}{2}: the integral equals 2\pi i\,f(z_0) = 2\pi i\sin\frac{\pi}{2} = 2\pi i.

If the circle had radius 1, the point would be outside and the integral would be 0 (Cauchy’s theorem).

The point pi over 2 inside the circle of radius 2
The contour encloses the singularity at pi over 2, marked with a red cross.

Exercise 3. One singularity inside and one outside

Compute \displaystyle\oint_{|z| = 2}\frac{z}{(z - 1)(z + 3)}\,dz.

Result: \frac{\pi i}{2}.

Solution

The singularities are 1 (inside) and -3 (outside). Write the integrand as \frac{f(z)}{z - 1} with f(z) = \frac{z}{z + 3}, holomorphic inside the circle. Integral formula: 2\pi i\,f(1) = 2\pi i\cdot\frac{1}{4} = \frac{\pi i}{2}.

With radius 4 both would be inside, and the result would be 2\pi i\left(\frac{1}{4} + \frac{3}{4}\right) = 2\pi i (the residues at 1 and -3 are \frac{1}{4} and \frac{-3}{-4} = \frac{3}{4}).

Singularities inside and outside the contour
The pole at 1, in red, is inside the circle of radius 2; the one at minus 3, in grey, is outside and does not contribute.

Intermediate level

Exercise 4. The derivative formula

Compute \displaystyle\oint_{|z| = 3}\frac{e^{2z}}{(z - 1)^3}\,dz.

Result: 4\pi i\,e^2.

Solution

Derivative formula with n = 2: \oint\frac{f(z)}{(z - z_0)^3}dz = \frac{2\pi i}{2!}f''(z_0), with f = e^{2z} and z_0 = 1 (inside).

f''(z) = 4e^{2z}, f''(1) = 4e^2. Integral: \pi i\cdot 4e^2 = 4\pi i\,e^2.

Typical mistake: forgetting the factorial, or using the first derivative with a cubed denominator. The power of the denominator is n + 1.

The triple pole at 1
The circle of radius 3 encloses the point 1, where the integrand has a pole of order 3.

Exercise 5. A circle not centred at the origin

Compute \displaystyle\oint_{|z - i| = 1}\frac{dz}{z^2 + 1}. What is the same integral on |z| = 2?

Result: \pi; on |z| = 2 it is 0.

Solution

z^2 + 1 = (z - i)(z + i). The circle with centre i and radius 1 contains i but not -i (which is at distance 2 from the centre). With f = \frac{1}{z + i}: 2\pi i\,f(i) = 2\pi i\cdot\frac{1}{2i} = \pi.

On |z| = 2 both are inside: residues \frac{1}{2i} and \frac{1}{-2i}, sum zero, integral 0. This could be predicted: for large |z|, \frac{1}{z^2 + 1} \approx \frac{1}{z^2}, and a large circle gives an integral tending to 0 by the ML estimate; by contour deformation, every circle containing both poles gives the same value, hence 0.

A circle centred at i
The contour encloses the pole at i, in red, and leaves out the one at minus i, in grey.

Exercise 6. An ML estimate

Show that, for R \gt 1, \left|\displaystyle\int_{C_R}\frac{dz}{z^2 + 1}\right| \le \dfrac{\pi R}{R^2 - 1}, where C_R is the upper semicircle of radius R, and deduce the limit as R \to \infty.

Result: the estimate follows from the reverse triangle inequality; the limit is 0.

Solution

On |z| = R: |z^2 + 1| \ge |z|^2 - 1 = R^2 - 1 (reverse triangle inequality |a + b| \ge |a| - |b|). So \left|\frac{1}{z^2 + 1}\right| \le \frac{1}{R^2 - 1} = M. The length of the semicircle is L = \pi R. ML estimate: \le\frac{\pi R}{R^2 - 1} \to 0.

This is the step that, in the in-depth study on residues, allows \int_{-\infty}^\infty\frac{dx}{x^2 + 1} = \pi to be computed by closing with the semicircle.

The semicircle C R
For large R the contribution of the arc tends to zero and only the integral on the real axis remains.

Exam level

Exercise 7. Cauchy’s estimate

A function f is holomorphic in the disc |z| \le 2 and |f(z)| \le 5 on the circle |z| = 2. How large can |f'''(0)| be at most? Is the bound attained?

Result: |f'''(0)| \le \frac{15}{4} = 3.75; yes, for f(z) = \frac{5}{8}z^3.

Solution

Cauchy’s estimate (in-depth study): |f^{(n)}(z_0)| \le \frac{n!\,M}{R^n} = \frac{6\cdot 5}{8} = 3.75.

It is attained by f = \frac{5}{8}z^3: on |z| = 2, |f| = \frac{5}{8}\cdot 8 = 5, and f''' = \frac{5}{8}\cdot 6 = 3.75.

Meaning: a holomorphic function bounded on a large disc cannot oscillate quickly at the centre. The derivatives are controlled by the values on the boundary.

Disc of given radius
The estimate uses the values of f on the circle to control the derivatives at the centre.

Exercise 8. A consequence of Liouville

Let f be holomorphic on the whole of \mathbb C with \text{Re}\,f(z) \le 0 for every z. Prove that f is constant.

Result: apply Liouville to g = e^{f}, which has modulus at most 1.

Solution

f is not necessarily bounded (the imaginary part can grow), so Liouville does not apply directly. Consider g(z) = e^{f(z)}: it is holomorphic on the whole of \mathbb C (composition of holomorphic functions) and |g| = e^{\text{Re}\,f} \le e^0 = 1. By Liouville g is constant.

Then g' = f'e^{f} = 0, and since e^{f} never vanishes, f' = 0 everywhere: f is constant.

This is the typical scheme: transform the function with another one (exponential, reciprocal, \frac{1}{f - c}) to obtain a bounded function.

Exercise 9. Maximum modulus

Find the maximum of |z^2 + 3z - 1| on the disc |z| \le 1 and the point where it is attained.

Result: \sqrt{13} \approx 3.61, at z = i (and at z = -i).

Solution

The polynomial is holomorphic: by the maximum modulus principle the maximum lies on the circle |z| = 1. With z = e^{it}, z^2 + 3z - 1 = e^{it}\left(e^{it} + 3 - e^{-it}\right) = e^{it}(3 + 2i\sin t). The modulus is |3 + 2i\sin t| = \sqrt{9 + 4\sin^2 t}, maximum for \sin t = \pm 1, i.e. z = \pm i: \sqrt{13}.

Check: at z = i, i^2 + 3i - 1 = -2 + 3i, modulus \sqrt{13}. At z = 1, the most “natural” value to try, one gets only 3.

The maximum point on the boundary of the disc
The maximum of the modulus is on the unit circle, at the point i marked in purple, and not inside the disc.

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