Answer to Question #102646 in Complex Analysis for BIVEK SAH

Question

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.

Answer 1

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

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

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

(2) $5(w+1)^2-3(w+1)+2=5(w^2+2w+1)-3w-3+2=5w^2+7w+4$

and

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

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 $w=0$ . 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 $w=r\cdot e^{i\phi}$ where $r,\phi \in R_1; r=const; \phi \in [0,2\pi]$. We rewrite integral (3) as

(4) $\oint_{\gamma} \frac{5w^2+7w+4}{w^3}dw=\oint_{\gamma} \frac{5}{w}dw+\oint_{\gamma} \frac{7}{w^2}dw+\oint_{\gamma} \frac{4}{w^3}dw=\\=5i \int^{2\pi}_{0} d\phi+7i \int^{2\pi}_{0} \frac{1}{re^{i\phi}}d\phi+4i \int^{2\pi}_{0} \frac{1}{(re^{i\phi})^2}d\phi$

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

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

$\int^{2\pi}_{0} \frac{1}{re^{i\phi}}d\phi =\frac{1}{r} \int^{2\pi}_{0} e^{-i\phi}d\phi=\frac{1}{r}(\frac{e^{-i\phi}}{-i})^{2\pi}_0=\frac{i}{r}(e^{-i2\pi}-e^{0})=\frac{i}{r}(1-1)=0$

The first one is

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

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

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