# Exercise sheet 9 pdf

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## Abstract

407sh9qu3.dvi Question Let Ds be the hyperbolic disc in the Poincare´ disc D with hyperbolic radius s, and let Cs be the hyperbolic circle with hyperbolic radius s that bounds Ds. Describe the behavior of the quotient q(s) = length D (Cs) areaD(Ds) as s→ 0 and as s→∞. Compare the behavior of q with the analogous quantity calculated using a Euclidean disc and a Euclidean circle. Answer We know from exercise sheet 8 that lengthD(Cs) = 2pi sinh(s). To calculate areaD(Ds): Recall that the euclidean radius of Ds is R = tanh( 1 2 s), and so areaD(Ds) = ∫ 2pi 0 ∫ R 0 4 (1− |z|2)2 dx dy = ∫ 2pi 0 ∫ R 0 4r dr dθ (1− r2)2 = 8pi ∫ R 0 r dr (1− r2)2 = 8pi 1 2 (1− r2) ∣∣∣∣ R 0 = 8pi ( 1 2(1−R2) − 1 2 ) = 4pi 1−R2 − 4pi = 4pi(1− 1 +R2) 1−R2 = 4pi tanh2(1 2 s) 1− tanh2(1 2 s) = 4pi sinh2( 1 2 s) 1 and so q(s) = length D (Cs) areaD(Ds) = 2pi sinh(s) 4pi sinh2(1 2 s) = 4pi sinh(1 2 s) cosh(1 2 s) 4pi sinh2(1 2 s) = cosh(1 2 s) sinh(1 2 2) = e 1 2 s + e −1 2 s e 1 2 s − e −1 2 s = es + 1 es − 1 as s→∞, q(s)→ 1. as s→ 0+, q(s)→∞ (since numerator → 2 and denominator → 0.) The analagous euclidean quantity is qE(s) = length D (Cs) areaD(Ds) = 2pis pis2 = 2 s as s→∞, qE(s)→ 0 (different from q(s)). as s→ 0+, qE(s)→∞ (as q(s)) 2

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