C \mathbb{C} omplex-icated Cauchy Contour Crescent Curve

Calculus Level 5

Let z z denote a complex number. Two sets of complex numbers -- one satisfying ∣ z ∣ = 1 |z| = 1 and another ∣ z + ∣ z ∣ ∣ = 1 |z + |z|| = 1 -- are illustrated in the Argand plane as shown on the right. If the area of the pink region can be expressed as

A B π − C D E \dfrac{A}{B}\pi - \dfrac{C}{D}\sqrt{E}

where A A , B B , C C , D D , and E E are positive integers, gcd ⁡ ( A , B ) = gcd ⁡ ( C , D ) = 1 \gcd(A,B) = \gcd(C,D) = 1 , and E E is square-free, input A + B + C + D + E A + B + C + D + E as your answer.

Bonus. For fun, try finding different ways of solving the problem.


The answer is 11.

Michael Huang by Michael Huang , 29, USA

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1 solution

Mark Hennings
Feb 5, 2021

Using polar coordinates, the noncircular curve is defined by 1 = ∣ r ( cos ⁡ θ + 1 ) + i r sin ⁡ θ ∣ = ∣ 2 r cos ⁡ 1 2 θ ( cos ⁡ 1 2 θ + i sin ⁡ 1 2 θ ) ∣ = 2 r cos ⁡ 1 2 θ 1=\big|r(\cos\theta+1)+i r\sin\theta\big| = \big|2r\cos\tfrac12\theta(\cos\tfrac12\theta+i\sin\tfrac12\theta)\big| \; = \; 2r\cos\tfrac12\theta so the curve is r = 1 2 sec ⁡ 1 2 θ ∣ θ ∣ < π r \; = \; \tfrac12\sec\tfrac12\theta \hspace{2cm} |\theta| < \pi Thus we want 1 2 ∫ − 2 3 π 2 3 π ( 1 − 1 4 sec ⁡ 2 1 2 θ ) d θ = 1 2 [ θ − 1 2 tan ⁡ 1 2 θ ] − 2 3 π 2 3 π = 2 3 π − 1 2 3 \tfrac12\int_{-\frac23\pi}^{\frac23\pi}\big(1 - \tfrac14\sec^2\tfrac12\theta\big)\,d\theta \; = \; \tfrac12\Big[ \theta - \tfrac12\tan\tfrac12\theta\Big]_{-\frac23\pi}^{\frac23\pi} \; = \; \tfrac23\pi - \tfrac12\sqrt{3} making the answer 2 + 3 + 1 + 2 + 3 = 11 2+3+1+2+3 = \boxed{11} .

3 Helpful 2 Interesting 4 Brilliant 0 Confused

If possible could you give me a little more detail on the integration part please, as I am not at all familiar with polar coordinate integration ( more on how you got the integral, as I understand how the integration was done )

Jason Gomez - 4 months, 1 week ago

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If we have a curve defined in polar coordiantes by the equation r = f ( θ ) r = f(\theta) , the area swept out between angles α \alpha and β \beta is 1 2 ∫ α β f ( θ ) 2 d θ \tfrac12\int_\alpha^\beta f(\theta)^2\,d\theta Check out this Wiki .

In this question, I want the area between the curves r = 1 r = 1 and r = 1 2 sec ⁡ 1 2 θ r = \tfrac12\sec\tfrac12\theta in the range − 2 3 π < θ < 2 3 π -\tfrac23\pi < \theta < \tfrac23\pi (the two curves intersect when θ = ± 2 3 π \theta = \pm\tfrac23\pi ). The rest is standard integration.

Mark Hennings - 4 months, 1 week ago

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Thanks a lot, just checking, this is like doing Riemann sums on sectors of a circle right, looks quite powerful

Jason Gomez - 4 months, 1 week ago

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@Jason Gomez – Yes, it can be thought of that way, where each δ θ \delta\theta element is approximated as a triangle.

Mark Hennings - 4 months, 1 week ago

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