Question #102646
Evaluate the integral

Integral [(5z^2-3z+2)]/[(z-1)^3] dz
Close curve C



where C is any simple closed curve enclosing z = 1.
1
Expert's answer
2020-02-19T10:44:11-0500

By the condition of the problem it is necessary to find the integral c5z23z+2(z1)3dz\oint_c \frac{5z^2-3z+2}{(z-1)^3}dz along the contour cc enclosing the point z=1. For simplicity we will make a replacement z=w+1z=w+1 then dz=dwdz=dw and integral become

(1) γ5(w+1)23(w+1)+2w3dw\oint_{\gamma} \frac{5(w+1)^2-3(w+1)+2}{w^3}dw

where γ\gamma is any contour enclosing point w=0w=0. Simplify numerator we have

(2) 5(w+1)23(w+1)+2=5(w2+2w+1)3w3+2=5w2+7w+45(w+1)^2-3(w+1)+2=5(w^2+2w+1)-3w-3+2=5w^2+7w+4

and

(3)γ5w2+7w+4w3dw\oint_{\gamma} \frac{5w^2+7w+4}{w^3}dw

Note that the integrand function is unambiguous, differentiable, and continuous in the entire area surrounded by the contour except for the point . This means that it is an analytic function in the specified area with the exception of point w=0w=0 . In the theory of analytic functions, it is proved that any change in the contour in the region of analyticity of a function does not change the value of the integral of this function along the contour. Then we can integrate about some circle, with w=reiϕw=r\cdot e^{i\phi} where r,ϕR1;r=const;ϕ[0,2π]r,\phi \in R_1; r=const; \phi \in [0,2\pi]. We rewrite integral (3) as

(4) γ5w2+7w+4w3dw=γ5wdw+γ7w2dw+γ4w3dw==5i02πdϕ+7i02π1reiϕdϕ+4i02π1(reiϕ)2dϕ\oint_{\gamma} \frac{5w^2+7w+4}{w^3}dw=\oint_{\gamma} \frac{5}{w}dw+\oint_{\gamma} \frac{7}{w^2}dw+\oint_{\gamma} \frac{4}{w^3}dw=\\=5i \int^{2\pi}_{0} d\phi+7i \int^{2\pi}_{0} \frac{1}{re^{i\phi}}d\phi+4i \int^{2\pi}_{0} \frac{1}{(re^{i\phi})^2}d\phi

We given that dw=rieiϕdϕdw=r\cdot i\cdot e^{i\phi}d\phi .

The second and third integrals in (4) equals 0. For example second integral is

02π1reiϕdϕ=1r02πeiϕdϕ=1r(eiϕi)02π=ir(ei2πe0)=ir(11)=0\int^{2\pi}_{0} \frac{1}{re^{i\phi}}d\phi =\frac{1}{r} \int^{2\pi}_{0} e^{-i\phi}d\phi=\frac{1}{r}(\frac{e^{-i\phi}}{-i})^{2\pi}_0=\frac{i}{r}(e^{-i2\pi}-e^{0})=\frac{i}{r}(1-1)=0

The first one is

5i02πdϕ=10πi5i \int^{2\pi}_{0} d\phi=10\pi i

Answer: The Integral c5z23z+2(z1)3dz=10πi\oint_c \frac{5z^2-3z+2}{(z-1)^3}dz=10\pi i

where C is any simple closed curve enclosing z = 1.



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