Residue theorem and integral of a complex-valued function

Question:
Find the value of
\[ \int_C \frac{dz}{z^2+2iz-4} \]
The contour, $C$, is a circle that has center $z=1$ and radius $\sqrt{2}$. It is directed positively.

Answer:
The integrand can be reduced in two factors.
\[ I = \int_C \frac{dz}{z^2+2iz-4} = \int_C \frac{dz}{(z+(\sqrt{3}+i))(z-(\sqrt{3}-i))} \]
The value when the denominator becomes zero is the singular point in the contour. $z=-(\sqrt{3}+i)$ and $z=\sqrt{3}-i$ can be the points, but only $z=\sqrt{3}-i$ is the pole inside contour.
Calculate the residue.
\[ \mathrm{Res}(f(z):\sqrt{3}-i) = \lim_{z\rightarrow \sqrt{3}-i}(z-(\sqrt{3}-i)f(z) \\
                           = \lim_{z\rightarrow \sqrt{3}-i}(z-(\sqrt{3}-i)\frac{1}{(z+(\sqrt{3}+i))(z-(\sqrt{3}-i))} \\
                           = \lim_{z\rightarrow \sqrt{3}-i}\frac{1}{z+(\sqrt{3}+i)}    \\
                           = \frac{1}{2\sqrt{3}}\]
Then, using the residue theorem, we have the value of integral:
\[ I = 2\pi i \times \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}\pi i}{3} \]

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