Mathematical Tools for Physics

14—Complex Variables 422

Just asξkis a point in thekthinterval, so isζka point in thekthinterval along the curveC.
How do you evaluate these integrals? Pretty much the same way that you evaluate line integrals in vector
calculus. You can write this as
∫

C

f(z)dz=

∫ (

u(x,y) +iv(x,y)

)(

dx+idy

)

=

∫ [

(udx−v dy) +i(udy+v dx)

]

If you have a parametric representation for the values ofx(t)andy(t)along the curve this is

∫t 2

t 1

[

(ux ̇−vy ̇) +i(uy ̇+vx ̇)

]

dt

For example take the functionf(z) =z and integrate it around a circle centered at the origin. x=Rcosθ,
y=Rsinθ. ∫

z dz=

∫

[

(xdx−y dy) +i(xdy+y dx)

]

=

∫ 2 π

0

dθR^2

[

(−cosθsinθ−sinθcosθ) +i(cos^2 θ−sin^2 θ)

]

= 0

Wouldn’t it be easier to do this in polar coordinates?z=reiθ.

∫

z dz=

∫

reiθ

[

eiθdr+ireiθdθ

]

=

∫ 2 π

0

ReiθiReiθdθ=iR^2

∫ 2 π

0

e^2 iθdθ= 0 (3)

Do the same thing for the function 1 /z. Use polar coordinates.

∮

1

z

dz=

∫ 2 π

0

1

Reiθ

iReiθdθ=

∫ 2 π

0

idθ= 2πi (4)

This is an important result! Do the same thing forznwherenis any positive or negative integer, problem 1.
Rather than spending time on more examples of integrals, I’ll jump to a different subject. The main results
about integrals will follow after that (the residue theorem).