A meromorphic function f has 3 zeros and 1 pole inside a closed contour γ. What is the winding number of the image curve f(γ) around the origin?
A3, counting only the zeros
B−1, counting only the pole with its sign
C2, since the argument principle gives Z − P = 3 − 1 = 2
D4, since the argument principle counts total singularities Z + P = 4
The argument principle states that (1/2πi) ∮_γ f'/f dz = Z − P, where Z counts zeros and P counts poles with multiplicity. Here Z − P = 3 − 1 = 2. The argument principle also equates this to the winding number of the image curve f(γ) around the origin. So the image curve wraps around the origin exactly 2 times. The winding number is a signed count of net wrapping—poles contribute negatively because f'/f has residue −m at a pole of order m.
Question 2 Multiple Choice
Near a zero of order n at z₀, the logarithmic derivative f'(z)/f(z) has:
AA pole of order n at z₀ with residue n
BA simple pole at z₀ with residue n
CA simple pole at z₀ with residue −n
DA zero of order n at z₀, mirroring the zero of f
Near a zero of order n, write f(z) = (z − z₀)ⁿ g(z) where g(z₀) ≠ 0. Then f'/f = n/(z − z₀) + g'/g. The term g'/g is analytic near z₀ (since g doesn't vanish there), so f'/f has a *simple* pole—not a pole of order n—at z₀, with residue exactly n (the order of the zero). This is why the residue theorem applied to f'/f counts zeros with their multiplicity. Near a pole of order m, the same calculation gives a simple pole with residue −m.
Question 3 True / False
The key reason the argument principle works is that f'/f transforms zeros and poles of f into simple poles with integer residues, making the residue theorem directly applicable to count them.
TTrue
FFalse
Answer: True
True. This is the mechanism that makes the argument principle work. If you tried to count zeros by integrating f itself, you would get no useful information—f is analytic near its zeros and the integral depends on global behavior. But f'/f converts every zero of order n into a simple pole with residue +n, and every pole of order m into a simple pole with residue −m. The residue theorem then gives the sum of these residues as (1/2πi)∮ f'/f dz = Z − P. The 'logarithmic derivative trick' is what makes zeros and poles into something the residue theorem can directly count.
Question 4 True / False
The argument principle counts zeros and poles by integrating f(z) itself around the contour, rather than its logarithmic derivative f'(z)/f(z).
TTrue
FFalse
Answer: False
False. The argument principle integrates the *logarithmic derivative* f'/f, not f itself. Integrating f around a closed contour would give zero for analytic f (by Cauchy's theorem) or information about residues of f at its poles—not a count of zeros. The key is that f'/f has simple poles at exactly the zeros AND poles of f, with integer residues that encode their orders. This is the mechanism that converts analytic information (where are the zeros?) into the language the residue theorem can answer.
Question 5 Short Answer
Explain in your own words why (1/2πi) ∮_γ f'/f dz equals the winding number of the image curve f(γ) around the origin. What is the geometric meaning of this identification?
Think about your answer, then reveal below.
Model answer: Since f'/f = d/dz [log f(z)], the integral ∮ f'/f dz measures the total change in log f as z traverses γ. Writing log f = log|f| + i·arg(f): because γ is closed, z returns to its starting point, so log|f| returns to its starting value and contributes zero net change. The only contribution is i times the total change in arg(f)—the total angle swept by the image f(z) around the origin. This total angle change divided by 2π is the winding number of f(γ) around 0. So Z − P = (total change in arg(f))/(2π) = winding number of f(γ) around the origin.
The geometric meaning is that you can count zeros minus poles *visually*: draw the image curve f(γ) and count how many times it wraps around the origin (counterclockwise positive, clockwise negative). Each zero of order n inside γ contributes n counterclockwise wraps; each pole of order m contributes m clockwise wraps. Rouché's theorem exploits this: if |f − g| < |f| on γ, the image curves of f and g are close enough to have the same winding number around the origin, so f and g have the same number of zeros inside γ. The topological picture (winding) and the analytic count (Z − P) are the same thing.