1: Prove that W 3:3 sin(:r) oo 374 + 15 using the Residue Theorem and showing all steps. You may as- sume the following inequality in your work: f0\" e'RSinmR dt 53 11' for any R 2: 1000. d3: = \"HE cosh/i] Question 2: Let P be a polynomial of degree n with distinct roots 21,...,zn and let Q be a polynomial of degree exactly a 2 (we can assmno here that n 23 2). Show that i Res(Q/P, at) = {l k=1 Question 4: Show that there is no entire function F that sat- ises the equation F2(z) = sin[z) for all complex numbers 2. Show further that the same conclusion holds if \" all complex numbers\" is replaced by \"all complex numbers in some open set\". Question 5: Determine how man},r zeroes he function ze" 1/4 has in the punctured disc A(U;0,2]. Do not attempt to nd the zeroes. 5. Find the integral of '3'\" rain :c L\" (as + 1)? \"'1' Solution: The required integral is the imaginary part of a: in: R in? IE _ me I = L (as + 1r \"I1 = shaman -3 (2:2 + 1)? \"I\" since the degree of the denominator is at least two more than the degree of the numerator. The integral is of type Ill[a}. So consider a oontour made up of a large semi-circle of radius R traversed in the anticlockwise direction [denoted by CR} and a straight line path from c: = R to c: 2 R along the :c-axis, and the complex valued function zen Hz} = m The onlyIr pole of the function inside the contour is at z = i. and ordsziz} = 2. Ev the residue theorem, R i: is 1-3 3!? . [.3 m r +1?\" M d\" ' 2\"'\""'\"=="\"}' letting R > no, the integral on the semicircle converges to zero, and so I = mugging) = negz g.\" i}2f[z} = 3:- So the integral that we want is the imaginary part of this, that is \"3 :rsinc: 1r d:c= . 1.\5. Find the integral of '3'\" rain :c L\" (as + 1)? \"'1' Solution: The required integral is the imaginary part of a: in: R in? IE _ me I = L (as + 1r \"I1 = shaman -3 (2:2 + 1)? \"I\" since the degree of the denominator is at least two more than the degree of the numerator. The integral is of type Ill[a}. So consider a oontour made up of a large semi-circle of radius R traversed in the anticlockwise direction [denoted by CR} and a straight line path from c: = R to c: 2 R along the :c-axis, and the complex valued function zen Hz} = m The onlyIr pole of the function inside the contour is at z = i. and ordsziz} = 2. Ev the residue theorem, R i: is 1-3 3!? . [.3 m r +1?\" M d\" ' 2\"'\""'\"=="\"}' letting R > no, the integral on the semicircle converges to zero, and so I = mugging) = negz g.\" i}2f[z} = 3:- So the integral that we want is the imaginary part of this, that is \"3 :rsinc: 1r d:c= . 1.\