Question

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, 5B() < M. [6] (b) A particle on the complex plane is trapped within a wall built along the unit circle. It travels from –i to e3ti/4 and then bouncing from e3ti/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] | / Blends V2+v2. (c) Evaluate the integral B(z)dz, 7 leaving your answer in the form of of a +ib for some real numbers a and b. (10)

Answer 1

Answer:- Given that itaz B(2) = 4-ziz By maximal modulus principle sup {13021} = Sup {1BC231} *B(z)) s sup{l Bcz)}} ZED 121-1 121=1 Now for Izlais 1+2.53 i +22 14-2 izl z lu+fzizl=6 u-ziz 2 So /Bcz) N it 22 4-212 2 1 / 2 so long | B(2)) << M = 1/2 L= distance between (0, -1) and 3tilu e 1 / 2 ) = 12 +12 1 IL By symmetry , length of Y=22 = 2√2+52 so | SB(z)dz) = 518cz)) dz 4 da 21 X 25245 rata 7 If B(?) dz| = 12 +52

2 s B(2) d2 dz 5+27 S+S St itzz dz u-ziz r2 3tilu y is the line joining i to Y₂ is the line joining e 3114 tox on Y, Z = X(t) tiy(t) X(t)=t -it Hati 9() = 15:15 (t) y (t) = ciet-iret So 2 = *(t)tiy (t) =t+i(-i-t-ist) (2= trimit to tuit dz=dt ti(-dt-irzat) dz= {(52+1)-1)dt itzz so S 11-212 dz = point (1-2t)i +((12+1) + +1) dt Y u-zit-zizt- 22dt SI