One loop that rebuilds a whole function

Mathematics · Complex Analysis · ages 20-21

Name ______________________ Date ____________

f is holomorphic on an open set containing the unit circle and its interior. The integral of f(z)/(z - 0.5) around it once counterclockwise is 10. What is the integral around the circle traversed twice counterclockwise?
Answer: ______________
f is holomorphic on an open set containing the unit circle and its interior. The integral of f(z)/(z - 0.5) around the circle once counterclockwise is 2πi f(0.5). What is it around the circle traversed once clockwise?
- 2πi f(0.5)
- -2πi f(0.5)
- 0
f is holomorphic on an open set containing the circle |z| = 2 and its interior. What is (1/(2πi)) times the loop integral of f(z)/(z - 1) around it once counterclockwise?
- f(1)
- f(2)
- 0
A function holomorphic on an open neighbourhood of a point has derivatives of all orders there. True or false?
Circle one: True False
The Cauchy estimate gives |f''(a)| ≤ 2! M/R². If f is holomorphic on a neighbourhood of the closed disk |z - a| ≤ 3 with |f| ≤ M on its boundary circle, and f''(a) = 4, what lower bound on M is guaranteed?
Answer: ______________
If |f(z)| ≤ 5 for every z in the complex plane and f is entire, then f is constant. True or false?
Circle one: True False
f is holomorphic on a neighbourhood of the closed disk |z - 1| ≤ 1, and f''(1) = 12. What is the loop integral of f(z)/(z - 1)³ around the boundary circle |z - 1| = 1 traversed once clockwise?
- -12πi
- 12πi
- 6πi
An entire function bounded on the unit disk must be constant. True or false?
Circle one: True False
f is holomorphic on a neighbourhood of the closed disk |z - a| ≤ 4, with |f| ≤ 6 on the circle |z - a| = 2 and |f| ≤ 20 on the circle |z - a| = 4. What is the tighter of the two second-derivative upper bounds on |f''(a)|?
Answer: ______________
Suppose a zero-free nonconstant polynomial p satisfies |p(z)| ≥ |z|³/2 for |z| ≥ 2 and |p(z)| ≥ 1/5 for |z| ≤ 2. What combined bound on |1/p| do these estimates give, and what follows?
- 5, so 1/p is bounded and entire, constant by Liouville, a contradiction
- 1/4, so 1/p is bounded and entire, constant by Liouville, a contradiction
- 5, so 1/p is unbounded and Liouville does not apply

Answer key

For grown-ups. Fold this page away before handing over the rest.

One loop that rebuilds a whole function W1-mt_mViFwrFNWQ-s1

20 · Repeated traversal multiplies by the winding number.
-2πi f(0.5) · Clockwise traversal reverses orientation and negates the integral.
f(1) · The formula returns the value at the interior point 1.
True · The uniform distance argument justifies repeated differentiation under the integral sign.
18 · 4 ≤ 2M/9 forces M ≥ 18.
True · One bound valid on the whole plane lets the Cauchy estimate force the derivative to 0 everywhere.
-12πi · The inverse formula gives 12πi counterclockwise, and clockwise traversal negates it to -12πi.
False · Boundedness on one disk is not a Liouville hypothesis; f(z) = z is a counterexample.
2.5 · Radius 2 gives 3 and radius 4 gives 2.5, so the tighter estimate is 2.5.
5, so 1/p is bounded and entire, constant by Liouville, a contradiction · Outside gives 1/4 and inside gives 5, so the combined bound is 5 and Liouville freezes the reciprocal.