Question

Let $f(z)=\sin z$ and consider the domain $D = \{z\in\mathbb{C}\mid 0< {Re \, z} < \pi , 0< {Im \, z} < 1\}$   (an open rectangle). Find the maximum of on $\overline{D} = ( D\cup \partial D)$ as well as the -value(s) at which attains this maximum value.

Answer 1

Question; het to) - Sinz and considere -the domain D-2260 {ox Rez < 11. OK Tm2 <19 an open rectangle, find the maximum of 1(2) 5 = (pudD) as well as the at which ifl attains this maximum value. on z values Ans:- To Modulus Principle, which solve this problem, We have to use Maximum says "B a function f12) is analytic inside and on the boundatry of the dosed Cuare C and f12) is not constant in the region then maximum modulus of f12) attained on the region, here the Region is boundary on the OL Rez LT, OL Im2 LI 2 Zte G Here I Sin Extiy) = (sinal= sine cos hğ + losa sinkry Sinx coshy + (- Sinx) Sinhty Sint Cashy - Cosbe Sinhiny Now to maximize Iszl in that region , we have to minimize CBR and maximize coshy, As ashy increases as y increases of cosx decreases with a He have to maximize y and minimize. Cox, there for the pont will be y=1, x=1 2. And I sh(+3) = (simtg ?(Sinh 1)"= 1 +(sinho)260) of f2) = sinz will be the value therefore the that is modulus z boundary point) will be the point on c and 1 | 12 | ! + (Sinh 1) will be the maximum value. (@+i)= =2