Problem sheet 6 Feb 15th 2005

MT290 Complex variable

Ex. 1 Evaluate the following integrals around C, the circle |z| = 1 taken anticlockwise, using Cauchy’s integral formula (or the formula for derivatives if appropriate).

(a) ∫_{C} \frac{cos z}{z} dz; (b) ∫_{C} \frac{e^{πz}}{(z − \frac{1}{4})^{2}(z − 4)} dz.

Ex. 2 Throughout this question |z| < 2. Let f(z) = \frac{1}{z} + g(z), where |g(z)| < A (a fixed constant). Let γ_{r} be the circle: re^{it}, 0 ≤ t ≤ 2π. Use the estimation lemma to show that for r < 2

|∫_{γ_{r}} f(z) dz − 2πi| ≤ 2πAr.

Deduce that

∫_{γ_{r}} f(z) dz ⟶ 2πi

as r → 0. [Note: there is nothing special about the |z| < 2 here, it’s there just to restrict our attention to a finite region.

Ex. 3 Show that |e^{iz}| ≤ 1 when Imz ≥ 0. Let

I_{R} = ∫_{γ_{R}} \frac{e^{iz}}{1 + z^{2}} dz,

where γ_{R} is the semicircle given by ϕ(t) = Re^{it}, 0 ≤ t ≤ π. Show that, for all large R,

|I_{R}| ≤ \frac{2π}{R} .

Hence deduce that I_{R} → 0 as R → ∞.

Ex. 4 (Compare our discussion of ∫_{γ} \frac{1}{z^{2}+4} dz.) Let f(z) be a non-constant polynomial of degree at least two, and C(R) a circle centred on 0 with radius R. Use the estimation lemma to show that

lim_{R→∞} ∫_{C(R)} \frac{1}{f(z)} dz = 0.

Hence show, using the deformation of contours theorem, that if all the roots of f(z) = 0 lie within the circle |z| = R, then

∫_{C(R)} \frac{1}{f(z)}dz = 0.

Ex. 5 Solve the quadratic equation x^{2} − 4x + 5 = 0. Hence write x^{2} − 4x + 5 = (x − α)(x − β). Use the semicircular contour γ(R): (−R, R) ∪ {Re^{it} : 0 ≤ t ≤ π} to evaluate the following two integrals:

∫_{−∞}^{∞} \frac{sin x}{x^{2} − 4x + 5} dx, ∫_{−∞}^{∞} \frac{cos x}{x^{2} − 4x + 5} dx.

Note: Workshops this week: Thursday 3pm: 325 Friday 12pm: Lec B From Feb. 24th onwards: Let’s see if Thursday 11am attracts more people... Thursday 11am: 325 Thursday 3pm: 325 But no Friday workshops any longer.