× Close
Join the Ques10 Community
Ques10 is a community of thousands of students, teachers, and academic experts, just like you.
Join them; it only takes a minute
Question: Evaluate $\int \frac{z - 1}{z^2 + 2z + 5}$ dz over the curve C
0

Where C is:

(i) |z|=1

(ii) |z+1+i|=2

(iii) |z+1-i|=2

Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

m4e(34) • 123 views
 modified 10 weeks ago  • written 5 months ago by
0

For poles, $z^2 + 2z + 5 = 0$

$z = \frac{-2 \pm \sqrt{2^2 - 4(1)(5)}}{2} \\ = \frac{-2 \pm \sqrt{-16}}{2} \\ = \frac{-2 \pm 4i}{2} = -1 \pm 2i \\ \therefore z = -1+2i,-1-2i$

Therefore, z = (-1,2) & z =(-1,-2) are the two poles.

Let, A(-1,2) and B(-1,-2)

(i) |z| = 1

c = (0,0) and r = 1

d(Ac) = $\sqrt{(0+1)^2 + (0-2)^2} = \sqrt{1+4} = \sqrt{5} \gt 1$

Therefore, A lies outside the circle.

d(Bc) = $\sqrt{(0+1)^2 + (0+2)^2} = \sqrt{1+4} = \sqrt{5} \gt 1$

Therefore,B lies outside the circle.

Both the poles lie outside the circle |z|=1

Therefore, by Cauchy's theorem, $\int_c \frac{z-1}{z^2 + 2z + 5} \,\, dz = 0$

(ii) |z+1+i| = 2

z+1+i=0, therefore, z= -1-i and r = 2

Therefore, centre = (-1,-1), r = 2

d(Ac) = $\sqrt{(-1+1)^2 + (2+1)^2} = \sqrt{3^2} = 3 \gt 2$

d(Bc) = $\sqrt{(-1+1)^2 + (-2+1)^2} = \sqrt{(-1)^2} = 1 \lt 2$

Therefore, A lies outside the circle while B lies inside the circle.

Therefore residue of f(z) at z = -1-2i is

$lim_{z \to (-1-2i)} (z+1+2i)[\frac{z-1}{z^2+2z+5}] \\ = lim_{z \to (-1-2i)} (z+1+2i)[\frac{z-1}{(z+1-2i)(z+1+2i)} \\ = \frac{-1-2i-1}{-1-2i+1-2i} = \frac{-2-2i}{-4i} = \frac{-1-i}{-2i} = \frac{1-i}{2}$

$\therefore \int_c \frac{z-1}{z^2+2z+5} \,\, dz = 2 \pi i[\frac{1-i}{2}] = \pi i(1-i)$

(iii) |z+1-i| = 2

For centre (z+1-i) = 0, z = -1+i

Therefore, centre = (-1,1), r = 2

Therefore, the poles are a(-1,2) and B(-1-2)

d(Ac) = $\sqrt{(-1+1)^2 + (1-2)^2} = \sqrt{(-1)^2} = 1 \lt 2$

Therefore, A lies inside the circle.

d(Bc) = $\sqrt{(-1+1)^2 + (-2-1)^2} = \sqrt{(-3)^2} = 3 \gt 2$

Therefore, B lies outside the circle.

Residue of f(z) at z = -1+2i is

$\lim_{z \to (-1+2i)} (z+1-2i)[\frac{z-1}{(z+1-2i)(z+1+2i)}] \\ = \frac{-1+2i-1}{-1+2i+1+2i} = \frac{-2+2i}{4i} = \frac{-2(1+i)}{-4} = \frac{1}{2}(1+i)$

$\therefore \int_c \frac{z-1}{z^2+2z+5} = 2 \pi i (\frac{1+i}{2}) = \pi i (1+i)$