Question

The poles of the function f(z) = sin z / cos z are at​

Answer 1

Answer:

The function has simple poles at

z

=

±

π

2

The singularity at

z

=

0

is not a pole - it is a removable singularity.

Explanation:

z=0

Answer 2

Answer: The Pole of the function f(z) = sin z / cos z is simple pole and it's singularity is at point $z= (2n+1)\frac{\pi }{2}$, where n ∈ R

Step-by-step explanation:

The poles of a function is defined as the points where the value of the denominator tends to zero or gives a singularity.

given function f(z) = sin z / cos z

put cos z = 0

z= (2n+1)$\frac{\pi}{2}$

where n ∈ R

when n =0 then z= $\frac{\pi}{2}$

for n = 1, $z=\frac{3\pi}{2}$

for n=2  $z = \frac{5\pi}{2}$ ...so on

Thus, for a given function f(z)= sin z/ cos z

the pole is of order 1 and is called simple pole and is given by $(2n+1)\frac{\pi }{2}$