$u$ and $v$ are real and imaginary part of a complex function $f(z)$ find $\oint udy+vdx$ over $C$

Question

Given that $u $ and $v$ be the real and imaginary part of complex function

$$\frac{1}{z^2-6z+8}$$ of complex variable $z=x+iy$

Let $C$ be the simple closed curve $|z|=3$ in counter clockwise then find the value of the

$$\oint_C u \mathrm{d}y+v\mathrm{d}x$$

solution i tried -Here the given equation $\displaystyle \frac{1}{z^2-6z+8}$ can be written as $$\frac{x^2-y^2+8}{(x^2-y^2+8)^2+(2xy-6x-6y)^2}+i\frac{2xy-6x-6y}{(x^2-y^2+8)^2+(2xy-6x-6y)^2}$$

where $$u=\frac{x^2-y^2+8}{(x^2-y^2+8)^2+(2xy-6x-6y)^2}\;\; and \;\;v=\frac{2xy-6x-6y}{(x^2-y^2+8)^2+(2xy-6x-6y)^2}$$

also we have to find the value of >$$\oint_C u \mathrm{d}y+v\mathrm{d}x$$ also from green's theorem >$$\oint_C u \mathrm{d}y+v\mathrm{d}x=\int\int_C(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y})dxdy$$

when i try of futher solve this the calculations become more and more complicated ,is there any short way or i am missing some hint here .

Please help

Thank you

Answer 1

Can we write $$ \oint f(z)\;dz = \oint (u+iv)(dx+idy) = \oint \left(u\;dx - v\;dy \right) + i \left(u\;dy+v\;dx\right) $$ so that we are computing the imaginary part of $\oint f(z)\;dz$. And that is easy using the residues of $f$.

Answer 2

I assume that you don't know the residue theorem which is the convenient way of dealing with this kind of problem.

Hint for a direct calculation. We have that $z^2-6z+8=(z-2)(z-4)$ and therefore $$\frac{1}{z^2-6z+8}=\frac{1/2}{z-2}-\frac{1/2}{z-4}.$$ Now split the integral and note that if $z=x+iy$ and $a\in \mathbb{R}$ then $$\frac{1}{z-a}=\frac{\overline{z}-a}{|z-a|^2}=\frac{(x-a)-iy}{(x-a)^2+y^2}.$$