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Question: How to know that the poles of \( \dfrac{\sin z}{z\cos z} \) at \( 0,\dfrac{\pi }{2},-\dfrac{\pi }{2}...

How to know that the poles of sin⁡zzcos⁡z\dfrac{\sin z}{z\cos z} at 0,π2,−π20,\dfrac{\pi }{2},-\dfrac{\pi }{2} are simple poles?

Explanation

Solution

Hint : We have to find the analytical value if the function has one in any of the give values of 0,π2,−π20,\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 sin⁡zzcos⁡z\dfrac{\sin z}{z\cos z} at the points 0,π2,−π20,\dfrac{\pi }{2},-\dfrac{\pi }{2} , we find if the poles are simple poles or not.
A function f(z)f\left( z \right) has a removable singularity at z=z0z={{z}_{0}} if f(z)f\left( z \right) is not defined at z=z0z={{z}_{0}} and defining a value for f(z)f\left( z \right) at z=z0z={{z}_{0}} makes it analytic.
We first check at z=0z=0 for f(z)=sin⁡zzcos⁡zf\left( z \right)=\dfrac{\sin z}{z\cos z} . We can see the function is not defined at z=0z=0 because of zz being in the denominator. The function can have a limit value at z=0z=0 .
lim⁡z→0 sin⁡zzcos⁡z=lim⁡z→0 sin⁡zz×lim⁡z→0 1cos⁡z=1\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=0z=0 and has a removable singularity. The point doesn’t have a simple pole.
With similar logic the function is singular at z=±π2z=\pm \dfrac{\pi }{2} . The value of cos⁡z\cos z vanishes there.
The limit of the function doesn’t exist there either.
However, the limit value of lim⁡z→π2 (z−π2)sin⁡zzcos⁡z\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.
lim⁡z→π2 (z−π2)sin⁡zzcos⁡z=lim⁡z→π2 sin⁡zz×lim⁡z→π2 z−π2cos⁡z=2πlim⁡z→π2 z−π2cos⁡z\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 lim⁡z→π2 z−π2cos⁡z\underset{z\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{z-\dfrac{\pi }{2}}{\cos z} is in the form of 00\dfrac{0}{0} form. So, we differentiate from.
lim⁡z→π2 z−π2cos⁡z=lim⁡z→π2 1−sin⁡z=−1\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,
lim⁡z→π2 (z−π2)sin⁡zzcos⁡z=−2π\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=π2z=\dfrac{\pi }{2} .
With a similar argument we can show that at point z=−π2z=-\dfrac{\pi }{2} , we can have a simple pole.
lim⁡z→π2 (z+π2)sin⁡zzcos⁡z =lim⁡z→π2 sin⁡zz×lim⁡z→π2 z+π2cos⁡z =2πlim⁡z→π2 z+π2cos⁡z=2πlim⁡z→π2 1−sin⁡z =−2π\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(z)=p(z)q(z)f\left( z \right)=\dfrac{p\left( z \right)}{q\left( z \right)} where p(z)p\left( z \right) are analytic at z0{{z}_{0}} and p(z0)=0p\left( {{z}_{0}} \right)=0 then f(z)=p(z)q(z)f\left( z \right)=\dfrac{p\left( z \right)}{q\left( z \right)} has a pole of order m if and only if q(z)q\left( z \right) has a zero of order m.