91Ó°ÊÓ

A singularity is a point where a function becomes undefined. In our exercise, we are dealing with the function \( f(z) = \frac{\cos z}{1-2 \sin z} \) at the point \( z = \frac{\pi}{6} \). To see if there's a singularity here, we need to examine where the denominator becomes zero. The denominator of our function, \( 1 - 2 \sin z \), should be evaluated at \( z = \frac{\pi}{6} \). Calculating \( \sin \frac{\pi}{6} \) gives us \( \frac{1}{2} \), making the denominator \( 1 - 2 \times \frac{1}{2} = 0 \). This tells us we have a singularity at that point because the function is undefined where the denominator is zero.

A simple pole is a particular type of singularity where the function's value approaches infinity in a linear manner. We identified a singularity in our function \( f(z) = \frac{\cos z}{1-2 \sin z} \) at \( z = \frac{\pi}{6} \), but we need to determine if it is a simple pole. Given that the numerator \( \cos z \) does not equal zero at \( z = \frac{\pi}{6} \), and the denominator is zero, this indicates we have a simple pole. This simplifies residue calculations significantly, which is a core characteristic of simple poles.

Residue calculation is crucial in complex analysis, especially when dealing with simple poles. The residue at a simple pole \( z = z_0 \) can be found using the formula: \( \text{Res}(f, z_0) = \frac{g(z_0)}{h'(z_0)} \). Here, \( f(z) = \frac{g(z)}{h(z)} \) where \( g(z) = \cos z \) and \( h(z) = 1 - 2 \sin z \) at \( z_0 = \frac{\pi}{6} \). First, calculate \( g(\frac{\pi}{6}) \) which is \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \). Then, find the derivative of \( h(z) \), which is \( h'(z) = -2 \cos z \), and evaluate it at \( \frac{\pi}{6} \) to get \( -2 \cos \frac{\pi}{6} = -\sqrt{3} \). Finally, the residue is \( \text{Res}(f, \frac{\pi}{6}) = \frac{ \frac{\sqrt{3}}{2} }{ -\sqrt{3} } = -\frac{1}{2} \).

The derivative of a function gives us important information about its behavior, especially near points of interest like singularities. For our function \( h(z) = 1 - 2 \sin z \), we need to find its derivative to aid in residue calculation at a simple pole. Using standard differentiation rules, \( h(z) \) differentiated with respect to \( z \) is \( h'(z) = -2 \cos z \). Evaluating this derivative at \( z = \frac{\pi}{6} \) is essential, yielding \( h'\left(\frac{\pi}{6}\right) = -2 \cos \frac{\pi}{6} = -\sqrt{3} \). This derivative value is then used in the final residue formula, showcasing the importance of accurately determining derivatives in complex function analysis.