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Cauchy-Goursat theorem

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
I quote a question from Yahoo! Answers

Compute the following line integral. ∫a (z^2+3z+4)dz, where a is the circle |z|=2 oriented c-clockwise?
I'm not sure what the circle |z|=2 means, so I can't parametrize the circle.
I have given a link to the topic there so the OP can see my response.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
The function $f(z)=z^2+3z+4$ is holomorphic on $D=\mathbb{C}$ (simply connected) and $\gamma\equiv|z|=2$ (circle with center at the origin and radius 2) is contained in $D$. According to the Cauchy-Goursat theorem, $\int_{\gamma}f(z)dz=0$. Without using the Cauchy-Goursat theorem: we have the differential form: $$w=f(z)dz=(u+iv)(dx+idy)=\ldots=P(x,y)dx+Q(x,y)dy$$ The parametric equations of $\gamma$ are $x=2\cos t,\;y=2\sin t$ with $t\in[0,2\pi]$. Then, $$\int_{\gamma}f(z)dz=\int_{\gamma}w=\int_0^{2\pi}P(x(t),y(t))x'(t)dt+Q(x(t),y(t))y'(t)dt=\ldots=0$$