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$f(z)= \dfrac{2z+3i}{z^{2}+1/4}$, $C$ is the unit circle centered at zero.

Though the answer is $12\pi$, but my answer is coming to be $4\pi i$ If we decompose it into partial fractions then $f(z)$ reduces to $\dfrac{4}{z-\frac{i}{2}} - \dfrac{2}{z+ \frac{i}{2}}$

Applying Cauchy Integral formula, the first part integrates to $8\pi i $ and the second part reduces to $4\pi i$. Thus my answer, but the actual answer is otherwise

Help appreciated, Soham

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Check your partial fractions. All quadratics factor over complex numbers. – user31373 Aug 14 '12 at 13:39
my bad, I am sorry, I was copy pasting in a hurry and forgot to cross check, the partial fractions are fixed – Soham Aug 14 '12 at 13:42
Maybe the textbook author was also copy pasting in a hurry. I don't see anything wrong with your solution. – user31373 Aug 14 '12 at 13:49
:) $[][][][][][][]$ – Soham Aug 14 '12 at 13:53
up vote 1 down vote accepted

$$f(z):=\frac{2z+3i}{z^2+\frac{1}{4}}=\frac{2z+3i}{\left(z-\frac{i}{2}\right)\left(z+\frac{i}{2}\right)}\Longrightarrow$$ $$\begin{align*}Res_{z=i/2}(f)=&\lim_{z\to i/2}\left(z-\frac{i}{2}\right)f(z)=\lim_{z\to i/2}\frac{2z+3i}{z+\frac{i}{2}}=4\\Res_{z=-i/2}=&\lim_{z\to -i/2}\left(z+\frac{i}{2}\right)f(z)=\lim_{z\to -i/2}\frac{2z+3i}{z-\frac{i}{2}}=-2\end{align*} $$ so we finally get $$\oint_C\frac{2z+3i}{z^2+\frac{1}{4}}dz=2\pi i(4-2)=4\pi i$$

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To not leave an unanswered question, I'll point out that it's easier to compute this integral through the residue at $\infty$. Indeed, there are no singularities outside of the unit circle. Unfortunately I always forget the sign conventions about residues at infinity, but I remember that any holomorphic function that is $o(|z|^{-1})$ integrates to zero around infinity. Here $\dfrac{2z+3i}{z^2+1/4}=\dfrac{2}{z}+o(|z|^{-1})$, hence the integral is $\displaystyle \int _{|z|=1}\frac{2}{z}\,dz =4\pi i$.

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