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 \(z=\infty\).