Mathematical Rules of Thumb, illustrated reader · Chapter 6

06Complex Analysis

Turn Analytic Structure into Bounds and Counts

4 demonstrations follow the chapter's rules. Choose a value, watch the figure and the numbers change, and check your prediction. Every choice is precomputed from the notebook calculations.

Ask the chapter skill

“Help me use Chapter 6 for my question. Choose a rule, check its assumptions, and show how the result changes when an input changes.”

Use math-thumb-complex-analysis from the companion's skill package. The demonstrations below also work on their own.

Examples use constructed inputs or the book's own values, disclosed in each panel. A picture illustrates a rule; its assumptions set its scope.

1Demonstration 1 of 4

See powers multiply complex arguments

Does a principal angle record every complete turn?

Raise z=e^{i·135°} to a power. The spiral tracks the unwrapped angle; the output arrow shows where the principal angle lands after wrapping.

(reiθ)m=rmeimθ (re^{i\theta})^m=r^m e^{im\theta}

Integer power m. Integer powers and a unit-modulus input at 135°. Principal arguments use a chosen branch.

Predict first. Does a principal angle record every complete turn?

Choose an example

See powers multiply complex arguments. Integer power 2 multiplies the argument: 135°×2=270°. The principal argument wraps this to -90°, dropping 1 full turn; only the unwrapped angle records them.
Integer power m: 2
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Input argument (degrees)
135
Unwrapped output argument
270
Principal output argument
-90
Full turns lost by principal value
1

Integer power 2 multiplies the argument: 135°×2=270°. The principal argument wraps this to -90°, dropping 1 full turn; only the unwrapped angle records them.

Use the idea

Use rule 6.1.1 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Integer powers and a unit-modulus input at 135°. Principal arguments use a chosen branch.

Check your understanding: Does a principal angle record every complete turn?
No. It wraps into its branch interval; the unwrapped argument and winding retain the turn count.

Book source: Rule 6.1.1: Use polar form for complex products and powers. Demonstration C06-D01. Worked illustration.

2Demonstration 2 of 4

Use an analytic disk to bound derivatives

Can you apply the disk bound to 1/z on a disk containing zero?

Increasing an available analytic radius strengthens derivative bounds when the same boundary modulus bound holds.

|f(k)(0)|≤k!MRk |f^{(k)}(0)|\leq\frac{k!M}{R^k}

Disk radius R. Analytic on and inside the closed disk; stipulated M=3. It cannot be applied across a pole.

Predict first. Can you apply the disk bound to 1/z on a disk containing zero?

Choose an example

Use an analytic disk to bound derivatives. If the function is analytic on and inside this disk and its boundary modulus is at most 3, |f″(0)|≤6/R²=1.5. Doubling the radius cuts this ceiling by four. A function that stays tame across a wide disk cannot bend sharply at its center.
Disk radius R: 2
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Disk radius R
2
Boundary bound M
3
Second derivative ceiling
1.5

If the function is analytic on and inside this disk and its boundary modulus is at most 3, |f″(0)|≤6/R²=1.5. Doubling the radius cuts this ceiling by four. A function that stays tame across a wide disk cannot bend sharply at its center.

Use the idea

Use rule 6.2.2 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Analytic on and inside the closed disk; stipulated M=3. It cannot be applied across a pole.

Check your understanding: Can you apply the disk bound to 1/z on a disk containing zero?
No. The function has a pole at zero, violating the analytic hypothesis.

Book source: Rule 6.2.2: Use Cauchy's estimate to bound derivative scale. Demonstration C06-D02. Worked illustration.

3Demonstration 3 of 4

Check strict boundary domination before counting zeros

What fails when a=1?

Rouché's theorem: if one term is strictly bigger than another everywhere on a circle, adding the smaller term cannot change how many zeros lie inside. Here z³ has size 1 on the circle, so for |a| below 1, z³+a keeps all three zeros inside.

f(z)=z3+a,|a|<|z3|=1 on |z|=1 f(z)=z^3+a,\quad |a|<|z^3|=1\text{ on }|z|=1

Constant a. Rouché requires strict domination on the contour; boundary zeros invalidate the simple interior count argument.

Predict first. What fails when a=1?

Choose an example

Check strict boundary domination before counting zeros. For a=1, strict boundary domination fails. At a=1 the zeros lie on the boundary; above one they lie outside. The failed theorem test must not be treated as a zero-count proof.
Constant a: 1
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Constant a
1
Strict domination |a| < 1
False
Zeros strictly inside disk
0

For a=1, strict boundary domination fails. At a=1 the zeros lie on the boundary; above one they lie outside. The failed theorem test must not be treated as a zero-count proof.

Use the idea

Use rule 6.3.3 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Rouché requires strict domination on the contour; boundary zeros invalidate the simple interior count argument.

Check your understanding: What fails when a=1?
All three roots lie on the unit circle and strict domination fails. The interior conclusion cannot be carried through unchanged.

Book source: Rule 6.3.3: Use Rouché when one boundary term dominates. Demonstration C06-D03. Worked illustration.

4Demonstration 4 of 4

Count only the poles the contour encloses

Why is the integral zero again at R=2, even though two poles are inside?

A residue is the single number that summarizes a pole. Grow the circle and watch which poles it swallows. The integral is 2πi times the residues inside, and nothing outside matters.

∮|z|=Rdz(z−0.5)(z−1.5)=2πi∑insideRes \oint_{|z|=R}\frac{dz}{(z-0.5)(z-1.5)}=2\pi i\sum_{\text{inside}}\operatorname{Res}

Contour radius R. Simple poles at 0.5 and 1.5 with residues −1 and +1; circles centred at 0 that never pass through a pole.

Predict first. Why is the integral zero again at R=2, even though two poles are inside?

Choose an example

Count only the poles the contour encloses. f(z)=1/((z−0.5)(z−1.5)) has residue −1 at 0.5 and +1 at 1.5. Only the pole at 0.5 is inside, so the integral is 2πi×(−1)≈-6.28319i. The pole outside has no effect.
Contour radius R: 1
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Contour radius
1
Poles enclosed
1
Sum of enclosed residues
-1
Predicted integral 2πi×sum
-6.28319i
Numerical integral
0-6.28319i

f(z)=1/((z−0.5)(z−1.5)) has residue −1 at 0.5 and +1 at 1.5. Only the pole at 0.5 is inside, so the integral is 2πi×(−1)≈-6.28319i. The pole outside has no effect.

Use the idea

Use rule 6.3.2 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Simple poles at 0.5 and 1.5 with residues −1 and +1; circles centred at 0 that never pass through a pole.

Check your understanding: Why is the integral zero again at R=2, even though two poles are inside?
The residues −1 and +1 cancel. Enclosing poles does not guarantee a nonzero integral; their residues must not sum to zero.

Book source: Rule 6.3.2: Turn a contour integral into a sum of enclosed residues. Demonstration C06-D04. Worked illustration.

Bring the idea to a question of your own

Choose the relationship that answers your question, check its conditions, and compare the result with the accuracy or decision threshold you need.

The chapter skill can adapt these calculations to your inputs. It should name the assumptions, explain what the result supports, and say what still needs evidence. The chapter workbook adds a lab and three exercises with answers.