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
Best Answer
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 $f$.