Brane Space: More Difficult Contour Integrals

Time for some more contour integrals. We have seen earlier examples of contour line integrals and now we look at a more detailed example:

We want to integrate around the closed contour for which f(z) = 3x + 2iy:

We check first to see if the function is analytic using the Cauchy –Riemann relations:

i) ¶ u/ ¶x = ¶ v/ ¶y and ii) ¶ u/ ¶y = - ¶ v/ ¶x

For f(z) = 3x + 2iy = u(x,y) + iv(x,y)

We have: ¶ u/ ¶x = 3 and ¶ v/ ¶y = 2, so ¶ u/ ¶x ¹ ¶ v/ ¶y

And:

¶ u/ ¶y = 0 = - ¶ v/ ¶x

Since condition (i) is not fulfilled the function is not analytic, hence we cannot evaluate using Cauchy’ theorem , i.e.

ò _C f(z) dz = 0

So we must integrate line segment by line segment, viz.

I = ò _C f(z) dz = ò _C (u + iv) (dx + idy) =

ò _C (3x + 2iy) (dx + idy)

= ò _C (3x dx - 2y dy) + i ò _C ( 2y dx + 3x dy)

I = å³_{n
= 1} [ò _C (3x dx - 2y dy) + iò _C ( 2y dx + 3x dy)

On C1: 0 < x < 1, y = 3, dy = 0

ò _C1 (3x dx - 2y dy) + i ò _C1 ( 2y dx + 3x dy)

= ∫ ₀ ¹ 3x dx + i∫ ₀ ¹ 6 dy = [3/2 x²] ₀ ¹ + i[6y]₀ ¹

= 3/2 + 6i

On C2: 3 < y < 5, x = 1, and dx = 0

ò _C2 (3x dx - 2y dy) + i ò _C2 ( 2y dx + 3x dy)

= ò₃ ⁵ ( -2y ) dy + i ò₃ ⁵ 3 dy = [- y²] ₃ ⁵ + i[3y]₃ ⁵

= - 16 + 6i

On C3: 1 < t < 0, x = t, and dx = dt, y = 2t +3, dy = 2dt

Then:

ò _C3 (3x dx - 2y dy) + i ò _C3 ( 2y dx + 3x dy)

= ∫ ₁ ⁰ [ 3t dt - 2(2t + 3) 2 dt] + i ∫ ₁ ⁰ [ 2(2t + 3) dt + 3t (2dt)]

= - ∫ ₁ ⁰ (5t + 12) dt + i∫ ₁ ⁰ (10t + 6) dt

= - [5t²/ 2 + 12t]₁ ⁰ + i[5t² + 6t]₁ ⁰= 29/2 – 11i

Then the contour integral value is:

I = ( 3/2 + 6i) + (-16 + 6i) + (29/2 – 11i) = 0 + i

Problems for Math Mavens:

1) For the closed path shown in the diagram below, evaluate the contour integral. I.

Let f(z) = z ²

2) For the example shown in the blog post (Fig. 1) , what would the value of I be if:

f(z) = 3x + i3y?

Brane Space