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710 CHAPTER 20 Complex Integration

1 f (z)
f (z 0 ) =u(x 0 , y 0 ) + iv(x 0 , y 0 ) = dz
2πi C z − z 0
iθ
1 2π f (z 0 +re )
iθ
= ire dθ
2πi 0 re iθ
1 2π
= u(x 0 +r cos(θ), y 0 +r sin(θ))dθ
2π 0
i 2π
+ v(x 0 +r cos(θ), y 0 +r sin(θ))dθ.
2π 0
By comparing the real part of the left and right sides of this equation, we have the conclusion of
the theorem.

If D is a bounded domain, then the set D consisting of D, together with all boundary points
of D, is a closed and bounded set in the x, y-plane. If u(x, y) is continuous on D, then u(x, y)
must achieve a maximum value on D. In general, this might occur at any point or number of
points of D. However, if u is also harmonic on D, then u(x, y) must achieve its maximum value
at a boundary point of D. A proof of this uses the fact that u has a harmonic conjugate v, enabling
us to work with a differentiable complex function f = u + iv.

THEOREM 20.6 The Maximum Principle

Let D be a bounded domain in the plane and suppose u is continuous on D and harmonic on D.
Then u(x, y) achieves its maximum value at a boundary point of D.

Proof Let v be a harmonic conjugate of u in D, and let f = u + iv. Define
g(z) = e f (z) .
Then g is differentiable on D.Now |g(z)| is a continuous function of two real variables on the
closed and bounded set D of points in the plane. By a theorem of calculus, |g(z)| achieves a
maximum value at a boundary point of D.But

e
|g(z)|=|e u(x,y)+iv(x,y) |=|e u(x,y) iv(x,y) |= e u(x,y) .
Since e u(x,y) is a strictly increasing function, e u(x,y) and u(x, y) must achieve their maximum
values at the same point. Therefore, u(x, y) must achieve its maximum at a boundary point.

20.3.5 Bounds on Derivatives

It is possible to bound the derivatives of a complex function in terms of a bound on the function.

THEOREM 20.7 Bounds on Derivatives

Suppose f is differentiable on an open set G, and z 0 is a point in G. Let the closed disk of radius
r about z 0 be entirely contained in G. Suppose that | f (z)|≤ M for z on the circle bounding this
disk. Then, for any positive integer n,
Mn!
| f (n) (z 0 )|≤ .
r n
Theorem 20.7 can be proved by parametrizing the circle bounding the disk as γ(t)=z 0 +re it
in Cauchy’s integral formula for f (n) (z 0 ).

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October 14, 2010 15:32 THM/NEIL Page-710 27410_20_ch20_p695-714