Some Solved problems on zeros and poles.

Few Useful results regaring the poles and zeros in terms of quotients and product of two functions,

• Let $f(z)=h(z)/g(z)$, where $h$ and $g$ are analytic in some open disk about $z_0$, then $f$ has a pole of order $n$ at $z_0$.
• Let $f(z)=h(z)/g(z)$ and suppose $h$ and $g$ are analytic in some open disk about $z_0$, with $n>k$, then $f$ has a pole of order $n-k$ at $z_0$.
• Let $f$ has a pole of order $m$ at $z_0$ and let $g$ has a pole of order $n$ at $z_0$. then $fg$ has a pole of order $m+n$ at $z_0$
• Given a function $f(z)$, we say that $f$ has a singularity (of a given type) at infinity if and only if $f(\frac1z)$ has a singularity (of said type) at $0$.

Here are some solved examples:

1. $f(z)=\frac{1+4z^3}{sin^6z}$ 

Solution:

at $z=n\pi$ denominator becomes zero. and numerator is not zero at any of these points.

so, $f(z)$ has pole of order $6$ at $n\pi$

2. $f(z)=\frac{(z-\frac{3\pi}{2})^4}{\cos^7z}$

Solution:

at $\frac{3\pi}{2}$ numerator has zero of order 4. and at this point denominator has zero of order $7$.

so, $f(z)$ has a pole of order $3$ at this point.
moreover, pole of order 7 at $n\pi/2$.(other than $3\pi/2$.)

3.  $f(z)=\frac{1}{z^2 sin z}$ 

solution:

at $z=0$, $\frac{1}{z^2}$ has pole of order 2 and $\frac{1}{\sin z}$ has a pole of order $1$ at $z=0$

so, $f(z)=\frac{1}{z^2 sin z}$ has a pole of order $3$ at $0$.
and simple pole at $z=n\pi$.

4. $f(z)=z^2-\frac{1}{z^2}$ 

Solution:

$f(z)=z^2-\frac{1}{z^2}=\frac{z^4-1}{z^2}$

at $z=0$, $f(z)$ has pole of order $2$.

at $z=1$ $f(z)$ has zero of order $1$.

5. $f(z)=\frac{1}{(z^2+a^2)^2}$.

Solution:

pole at  $z=\pm a i$ of order $2$.

6. $f(z)=\frac{\sin^2 z}{z^2}$ 

Solution:

numerator:

at $z=0$ zero of order $2$

Denominator:

At $z=0$, zero of order $2$.

So, at $z=0$, $f(z)$, has no singularity.

and zero of order $2$ at $(2n+1)\pi$.

7. $f(z)=(z+1)\sin(\frac{1}{z-2})$

Solution:

$=(z+1)(\frac{1}{z-2})+\frac{1}{z-2}^3\times \frac{1}{3!}+\cdots$

$=\frac{z+1}{z-2}+\frac{z+1}{(z-2)^3}\frac{1}{3!}+\cdots$

essential singularity at $z=2$

zero at $z=-1$

8. $f(z)=\frac{e^{2z}}{(z-1)^4}$

Solution:

numerator is not zero for any $z$.

and denominator has zero of order 4 at $1$.

so $f(z)$ has pole of order at $z=0$ of order $4$.

9. $f(z)=\cos^2(\frac{z}{2})$

Solution:

zero at $z=n\pi$ of order 2.

10. $f(z)=(z^2+1)(e^z-1)$

solution:

at $z=0$  zero of order $1$.

at $z=\pm i$ zero of order $1$

11. $f(z)=(z^4-z^2-6)^3$

Solution:

$Z^4-Z^2-6=0 \Rightarrow z=3,-2$

at $z=-2,3$ zero of order $3$.

12. $f(z)=\frac{z^4}{\sin z}$ 

Solution:

Numerator: zero of order $4$ at $z=4$

Denominator: zero of order $1$ at $n\pi$

So, at $z=0$ zero of order $3$

and at $z=n\pi$ pole of order $1$.

13. $f(z)=\sin \frac{1}{z}$

Solution:

essential singularity at $z=0$

14. $f(z)=\frac{1-\cot z}{z}$

Solution:

zero of order $1$ at $2\pi +\frac{\pi}{4}$

and pole of order $1$ at $0$

15. $f(z)=cos^3 z$

Solution: 

zero of order $3$ at $(2n+1)\frac{\pi}{2}$

16. $f(z)= \frac{(\pi-2)(z^4-3z^2)}{\sin^2 z}$

Solution:

At $z=0$ numerator has zero of order $3$

and denominator has zero of order $2$

so, $f(z)$ has zero of order 1 at this point.

At $z=\pi)$  numerator has zero of order $1$

and denominator has zero of order $2$

so, $f(z)$ has pole of order 1 at this point.

now let look $1/f(z)$

$1/f(z)=\frac{\sin^2z}{(\pi-z)(z^4-z^2)}$
and this function has essential singularity at $z=0$.

so, $f(z)$ has essential singularity at $z=\infty$.