Just to check, would this be right?
$$\int_{|z|=1} \frac {1}{8(z-\frac{1}{2})^3} dz = 2\pi i f''(1/2) = 0$$
Where $f(z) = \frac{1}{8}$ by CIF? In the solutions it states $f(z) =1$; or does that not matter, they're both analytic aren't they? Thanks!
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2That integrand is a derivative, so integration about a loop is zero. Note the fact that $\left( (z - 1/2)^{-2}\right)' = -2(z-1/2)^{-3}$. – ncmathsadist May 10 '13 at 11:15
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Just note that, the integrand has a pole at $z=1/2$ of order $3$. See here for how to find the residue. – Mhenni Benghorbal May 10 '13 at 13:35
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Consider $ f(z) = \frac{1}{8}$, then by Cauchy Integral formula $$f^{(2)}(1/2) = 0 = \frac{2!}{2 \pi i} \oint_{|z| = 1} \frac{1}{8(z-1/2)^3}dz$$
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