Question #236289

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

Explain the problem with step by step process?

Expert's answer

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

Hence

"\\oint_Cz^3\\cos(\\frac{1}{z})dz = \\int_0^2 z^3\\cos(\\frac{1}{z})dz\\\\\n\\int \\:uv'=uv-\\int \\:u'v\\\\\nLet \\space u = z^3 \\implies du = 3z^2dz; dv= \\cos (\\frac{1}{z})dz \\implies v=z\\cos \\left(\\frac{1}{z}\\right)+\\text{sin}\\left(\\frac{1}{z}\\right)\\\\\n\\implies z^4\\cos \\left(\\frac{1}{z}\\right)+\\text{sin}\\left(\\frac{1}{z}\\right)- \\int 3z^3\\cos \\left(\\frac{1}{z}\\right)+\\text{sin}\\left(\\frac{1}{z}\\right)\\\\\n\\implies z^4\\cos \\left(\\frac{1}{z}\\right)+\\text{sin}\\left(\\frac{1}{z}\\right)- 3z^3\\int \\cos \\left(\\frac{1}{z}\\right)+\\int \\text{sin}\\left(\\frac{1}{z}\\right)\\\\\n\\implies [z^4\\cos \\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[2^4\\cos \\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)]\\\\-[0^4\\cos \\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"

Learn more about our help with Assignments: Chemical Engineering

## Comments

## Leave a comment