Question

How to know that the poles of $ \dfrac{\sin z}{z\cos z} $ at $ 0,\dfrac{\pi }{2},-\dfrac{\pi }{2} $ are simple poles?

Answer 1

Hint: We have to find the analytical value if the function has one in any of the give values of $ 0,\dfrac{\pi }{2},-\dfrac{\pi }{2} $ . Then we try to find the limit value of the function at those points to find the numerator value being non-zero.

Complete step by step solution:
From the singularity of the given function $ \dfrac{\sin z}{z\cos z} $ at the points $ 0,\dfrac{\pi }{2},-\dfrac{\pi }{2} $ , we find if the poles are simple poles or not.
A function $ f\left( z \right) $ has a removable singularity at $ z={{z}_{0}} $ if $ f\left( z \right) $ is not defined at $ z={{z}_{0}} $ and defining a value for $ f\left( z \right) $ at $ z={{z}_{0}} $ makes it analytic.
We first check at $ z=0 $ for $ f\left( z \right)=\dfrac{\sin z}{z\cos z} $ . We can see the function is not defined at $ z=0 $ because of $ z $ being in the denominator. The function can have a limit value at $ z=0 $ .
 $ \underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{z\cos z}=\underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{z}\times \underset{z\to 0}{\mathop{\lim }}\,\dfrac{1}{\cos z}=1 $ .
Therefore, the function is analytic at $ z=0 $ and has a removable singularity. The point doesn’t have a simple pole.
With similar logic the function is singular at $ z=\pm \dfrac{\pi }{2} $ . The value of $ \cos z $ vanishes there.
The limit of the function doesn’t exist there either.
However, the limit value of $ \underset{z\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( z-\dfrac{\pi }{2} \right)\dfrac{\sin z}{z\cos z} $ exists. We find the limit value.
 $ \underset{z\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( z-\dfrac{\pi }{2} \right)\dfrac{\sin z}{z\cos z}=\underset{z\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{\sin z}{z}\times \underset{z\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{z-\dfrac{\pi }{2}}{\cos z}=\dfrac{2}{\pi }\underset{z\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{z-\dfrac{\pi }{2}}{\cos z} $ .
The limit $ \underset{z\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{z-\dfrac{\pi }{2}}{\cos z} $ is in the form of $ \dfrac{0}{0} $ form. So, we differentiate from.
 $ \underset{z\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{z-\dfrac{\pi }{2}}{\cos z}=\underset{z\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{1}{-\sin z}=-1 $ . Therefore,
$ \underset{z\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( z-\dfrac{\pi }{2} \right)\dfrac{\sin z}{z\cos z}=-\dfrac{2}{\pi } $ .
Therefore, it has a simple pole at $ z=\dfrac{\pi }{2} $ .
With a similar argument we can show that at point $ z=-\dfrac{\pi }{2} $ , we can have a simple pole.
 $ \underset{z\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( z+\dfrac{\pi }{2} \right)\dfrac{\sin z}{z\cos z}\\
=\underset{z\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{\sin z}{z}\times \underset{z\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{z+\dfrac{\pi }{2}}{\cos z}\\
=\dfrac{2}{\pi }\underset{z\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{z+\dfrac{\pi }{2}}{\cos z}=\dfrac{2}{\pi }\underset{z\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{1}{-\sin z}\\
=-\dfrac{2}{\pi } $ .

Note: The pole is dependent on the numerator value. we have a quotient function $ f\left( z \right)=\dfrac{p\left( z \right)}{q\left( z \right)} $ where $ p\left( z \right) $ are analytic at $ {{z}_{0}} $ and $ p\left( {{z}_{0}} \right)=0 $ then $ f\left( z \right)=\dfrac{p\left( z \right)}{q\left( z \right)} $ has a pole of order m if and only if $ q\left( z \right) $ has a zero of order m.