The Argument Principle and Rouche's Theorem · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
به انگلیسی

Count roots by watching winding

Mathematics · Complex Analysis · ages 21-22
Name ______________________   Date ____________

  1. For f(z) = (z-1)/(z-2)^2 on a contour enclosing both 1 and 2, what is N minus P?

    Answer: ______________

  2. A closed curve has image f(C) that winds once clockwise around 0. What is the winding number?

    • -1
    • 1
    • 0

  3. Let f be meromorphic on a neighbourhood of a once-CCW simple closed piecewise C1 curve and its filled interior, with no zeros or poles on the curve. What does (1/(2 pi i)) times the loop integral of f prime over f equal?

    • Zeros minus poles inside, counted with multiplicity
    • Zeros plus poles inside, counted with multiplicity
    • The degree of f on the whole plane

  4. In the argument principle, zeros and poles count with multiplicity.

    Circle one:   True   False

  5. For f(z) = (z-1)(z+1) / (z-2) on the circle |z| = 3 traversed once counterclockwise, what is the winding number of the image f(C) around 0?

    Answer: ______________

  6. A nonconstant holomorphic function f on a connected open domain D maps open sets to what, and why?

    • Closed sets, since bounded images are bounded
    • Single points, since f is locally constant
    • Open sets, since nearby values are attained

  7. For f(z) = (z-1)^3 (z+2) on a contour enclosing both 1 and -2, what is N?

    • 3
    • 4
    • 5

  8. A nonconstant holomorphic function on a bounded domain that extends continuously to the closure must attain its maximum modulus on the boundary.

    Circle one:   True   False

  9. For p(z) = z^4 + 3z + 1, use Rouche to find the zero counts inside |z| = 1 and |z| = 2, then determine the number in the annulus 1 < |z| < 2.

    • 1
    • 3
    • 4

  10. For p(z) = (z-1)^2 (z-4) + 1 on |z-1| = 2, take the dominant group f(z) = (z-1)^2 (z-4). Find a lower bound for |f|, verify |1| < |f|, and count the zeros of p inside |z-1| = 2.

    • 1
    • 2
    • 3
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Answer key

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

Count roots by watching winding W1-mt_J1dC428M84-s1

  1. -1 · Zero at 1 has order 1, pole at 2 has order 2, so N-P = -1.
  2. -1 · Clockwise winding is negative; one clockwise turn gives -1.
  3. Zeros minus poles inside, counted with multiplicity · Each zero adds its multiplicity and each pole subtracts its own.
  4. True · A double zero winds twice and contributes two to the count.
  5. 1 · Zeros at -1 and 1 give N=2, pole at 2 gives P=1, so winding = N-P = 1.
  6. Open sets, since nearby values are attained · Isolated zeros and Rouche show every nearby value is attained, so the image is open.
  7. 4 · Zero at 1 has order 3, zero at -2 has order 1, so N = 4.
  8. True · Maximum modulus forbids an interior peak; continuity on the compact closure forces a boundary maximum.
  9. 3 · On |z|=1, |3z|=3 > 2 ≥ |z^4+1| gives 1 zero. On |z|=2, |z^4|=16 > 7 ≥ |3z+1| gives 4 zeros. Annulus: 4-1=3.
  10. 2 · |(z-1)^2(z-4)| ≥ 4 > 1 on |z-1|=2. Group has 2 zeros inside (at 1) and 1 outside (at 4).
Worksheet · LightMySky