# Answer in Chemical Engineering for pavani #231380

5) Evaluate ∮c z³ Cos 1/z Where C is |z|=2.

“|z| =2” means that the limits of integration are from 0 to 2

Hence

“oint_Cz^3cos(frac{1}{z})dz = int_0^2 z^3cos(frac{1}{z})dz\nint :uv’=uv-int :u’v\nLet space u = z^3 implies du = 3z^2dz; dv= cos (frac{1}{z})dz implies v=zcos left(frac{1}{z}right)+text{sin}left(frac{1}{z}right)\nimplies z^4cos left(frac{1}{z}right)+text{sin}left(frac{1}{z}right)- int 3z^3cos left(frac{1}{z}right)+text{sin}left(frac{1}{z}right)\nimplies z^4cos left(frac{1}{z}right)+text{sin}left(frac{1}{z}right)- 3z^3int cos left(frac{1}{z}right)+int text{sin}left(frac{1}{z}right)\nimplies [z^4cos left(frac{1}{z}right)+text{sin}left(frac{1}{z}right)- 3z^3 sin left(frac{1}{z}right)-frac{1}{z^2}text{cos}left(frac{1}{z}right)]|_0^2\n[2^4cos left(frac{1}{2}right)+text{sin}left(frac{1}{2}right)- 3*2^3 sin left(frac{1}{2}right)-frac{1}{2^2}text{cos}left(frac{1}{2}right)]\-[0^4cos left(frac{1}{0}right)+text{sin}left(frac{1}{0}right)- 3*0^3 sin left(frac{1}{0}right)-frac{1}{0^2}text{cos}left(frac{1}{0}right)]\n=15.6534”