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I need to integrate a complex function through the curve $\vert z+i\vert = 1$. As far as I know I need the parametric form of this curve. I know that when I have $\vert z\vert = 1$, the parametric form is something like $\cos(t) + i\sin(t)$. But what's different when I have that "$+i$"?

gebruiker
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Gustavo Amgarten
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  • You probably don't need a parametrisation, more likely Cauchy's integral theorem or integral formula (or another form of the residue theorem). But if you have a parametrisation of one curve, it's easy to get the parametrisation of translations of that curve, just add the appropriate constant. Here $-i + e^{it} = cos t + i((\sin t) - 1)$. – Daniel Fischer Jun 26 '14 at 13:32
  • Hint: the center of your circle has now been translated to $-i$. – Marc Jun 26 '14 at 13:32

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$$ |z-z_0| = r $$ is the equation of a circle centered in $z_0$ with radius r. Its parametric form is $$ z = z_0 + re^{it} = z_0 + r(\cos t + i \sin t) $$

Emanuele Paolini
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