Question

Problem 7.1. Let f(z) z²+2 (z²+3) (z² +2z+1) and let CR denote the semicircle of radius R parameterised by z(t) limf(z) dz = 0. JCR Reit with t€ [0, π]. Show that

Answer 1

Let's use the Cauchy Residue Theorem to solve this problem.

First, we need to find the poles of f(z).

f(z) has poles at z = i, -i, -1.

Let's consider the semicircle CR of radius R with center at the origin, parameterized by z(t) = Re^(it) with t in [0, π].

Using the Cauchy Residue Theorem, we have:

∫CR f(z) dz = 2πi Res(f, i) + 2πi Res(f, -i) + 2πi Res(f, -1)

Let's find the residues of f(z) at each of the poles.

At z = i, we have:

Res(f, i) = lim(z→i) (z-i) f(z)

= lim(z→i) (z-i) [(z²+2) (z²+3) (z²+2z+1)]

= (i²+2) (i²+3) (i²+2i+1)

= -4i

At z = -i, we have:

Res(f, -i) = lim(z→-i) (z+i) f(z)

= lim(z→-i) (z+i) [(z²+2) (z²+3) (z²+2z+1)]

= (-i²+2) (-i²+3) (-i²-2i+1)

= 4i

At z = -1, we have:

Res(f, -1) = lim(z→-1) (z+1) f(z)

= lim(z→-1) (z+1) [(z²+2) (z²+3) (z²+2z+1)]

= (1²+2) (1²+3) (1²-2+1)

= 60

Substituting these values into the Cauchy Residue Theorem, we get:

∫CR f(z) dz = 2πi (-4i) + 2πi (4i) + 2πi (60)

= 120π

Therefore, lim(R→∞) ∫CR f(z) dz = 120π.