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 denominator of integrand can be factored as follows:

\[ I = \int_C \frac{dz}{z^2+2iz-4} = \int_C \frac{dz}{(z+(\sqrt{3}+i))(z-(\sqrt{3}-i))} \]

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 singularities, but only \(z=\sqrt{3}-i\) becomes the pole inside the 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:

\[ I = 2\pi i \times \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}\pi i}{3} \]