One loop that rebuilds a whole function

Mathematics · Complex Analysis · ages 20-21

Name ______________________ Date ____________

The standard positive parametrisation of the unit circle winds around the origin how many times?
Answer: ______________
Let f be analytic inside and on the circle |z| equal 2. What is 1 divided by (2 pi i) times the loop integral of f(z) divided by (z minus 1) dz?
- f(2)
- 0
- f prime of 1
- f(1)
f is analytic inside and on the circle of radius 2. What is 1 over 2 pi i times the loop of f over (z minus 1)?
- f of 1
- f of 2
- 0
One complex derivative guarantees derivatives of all orders.
Circle one: True False
What is the second derivative of z cubed at the origin?
Answer: ______________
Complex sine is a bounded nonconstant entire function.
Circle one: True False
Let f be analytic inside and on the unit circle. What is the integral of f(z) divided by (z minus 0.5) dz around it?
- 2 pi i times f(0.5)
- 0
- 2 pi i times f(0)
- f(0.5)
A bounded entire function need not be constant: complex sine is a bounded counterexample. True or false?
Circle one: True False
The derivative estimate bounds f prime by M over R on a circle of radius R. With M equal 5 and R equal 20, what is the bound?
Answer: ______________
How does Liouville's theorem give the fundamental theorem of algebra?
- Every polynomial is a bounded function
- Liouville's theorem counts the zeros directly
- A nonconstant polynomial without zeros would have a bounded reciprocal, hence constant, a contradiction
- Polynomials grow too slowly at infinity to escape

Answer key

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

1 · One anticlockwise turn gives winding number 1.
f(1) · The formula returns the value at the interior point 1, namely f(1).
f of 1 · The formula returns the value at the interior point 1.
True · The integral formula differentiates under the sign any number of times.
0 · The second derivative is 6z, which vanishes at the origin.
False · Liouville forces constancy, and complex sine grows on the imaginary axis.
2 pi i times f(0.5) · The point 0.5 is interior, so the formula gives 2 pi i times f(0.5).
False · False. Liouville forces constancy, and complex sine is unbounded along the imaginary axis.
0.25 · 5 divided by 20 equals 0.25, and letting R grow forces the derivative to 0.
A nonconstant polynomial without zeros would have a bounded reciprocal, hence constant, a contradiction · No zeros makes 1 divided by p entire; growth control bounds it; Liouville freezes it; contradiction.