MHB Compute Line Integral with Cauchy-Goursat Theorem

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The discussion centers on computing the line integral ∫a (z^2+3z+4)dz over the circle |z|=2, oriented counterclockwise. The function f(z) = z^2 + 3z + 4 is holomorphic in the complex plane, allowing the application of the Cauchy-Goursat theorem, which states that the integral over a closed curve in a simply connected domain is zero. The parametric equations for the circle are provided, confirming that the integral evaluates to zero without needing the theorem. The conclusion emphasizes that both methods yield the same result, reinforcing the theorem's validity in this context.
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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.
 
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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, $$\int_{\gamma}f(z)dz=\int_{\gamma}w=\int_0^{2\pi}P(x(t),y(t))x'(t)dt+Q(x(t),y(t))y'(t)dt=\ldots=0$$
 
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