Question

6. Let B(2) i + 22 4- 2iz (a) Find the smallest positive real value M such that for every z on the closed unit disk D, B(2) <M. [6] (b) A particle on the complex plane is trapped within a wall built along the unit circle. It travels from -i to e3ri/4 and then bouncing from e3mi/4 to 1. Denote by y the curve representing the trajectory of the particle. Without evaluating the integral, show how we can obtain the following estimate. [6] |/ BCE) B(z)dz < V2+VZ. (c) Evaluate the integral B(z)dz, leaving your answer in the form of of a +ib for some real numbers a and b. [10]

Answer 1

Giren that Bcz) = it 2 Z 4-2iZ real we need to find the Smallest Positive Valile Apply Manimal modulus principle. Sup{ 1B17)% = sup{ IBC 2013 LED 121=1 $ | B (2) 1 & Supę 1 B (2) 14 2-) and 4-2121 < b) for 121=1, litzle H2=3 14-2 12 1> 14+ 12121) =6 SO |BC2) 1 = itz M = 1/2 particle travelled from ito and bounce back from eenila to 11 La distance blw (0, -15 CM2)ta): N2tre e 3711/4 31114 lengthg T=2L so | $ BC23d2= $ 10(2)|d25$$42: 22 2 • √2+12 CS Scanned with CamScanner

C) Evaluate integral 2 B(27 dz. P+2 z dz + 4-212 7 4-212 - 2 riline joining -i to e3 (114 127 8391114 Es on mi) Z= n(t) diy (t) s(t)=t, ylt)= -it 12+ ict) I setso arot-iſt t- tt Jatt 2t & dt = dt til-dt-iredt) (Cu2+1)-1)dt it 2 z dz (1-2 t)i+ Costut+1) dt 4-2iZ 4-lit-21-24-22 it so so s CS Scanned with CamScanner