# Singularities of Complex Functions

#### timeforchg

##### New member
Determine the location and nature of singularities in the finite z plane of the following functions:
(a) f(z) = (
- 1) sin(z)/[z(z+1)(z+2)(z-3)]
(b) g(z) = [1 + cos(z)]/

Using Cauchy's intergral formulae, referring to the above functions,
Evaluate
i)
f(z) dz, with C : | z + j | = 4 , traversed positively (CCW),
ii)
g(z) dz, with C: | z - 1 | = 2, traversed positively (CCW).

In each case, sketch the required contour C, carefully showing its direction.

#### Sudharaka

##### Well-known member
MHB Math Helper
Determine the location and nature of singularities in the finite z plane of the following functions:
(a) f(z) = (
- 1) sin(z)/[z(z+1)(z+2)(z-3)]
(b) g(z) = [1 + cos(z)]/

Using Cauchy's intergral formulae, referring to the above functions,
Evaluate
i)
f(z) dz, with C : | z + j | = 4 , traversed positively (CCW),
ii)
g(z) dz, with C: | z - 1 | = 2, traversed positively (CCW).

In each case, sketch the required contour C, carefully showing its direction.
Hi timeforchg, Let me help you with the second function. It is clear that $$z=0$$ is the only singular point (since $$g$$ is not defined only at $$z=0$$). Now,

$g(z)=\frac{h(z)}{(z-0)^8}\mbox{ where }h(z)=1+\cos z$

Note that the function $$h$$ is holomorphic everywhere(entire function). Therefore $$z=0$$ is a non-essential singularity (pole) of order 8.

$\oint_C g(z)\,dz=\oint_C\frac{1+\cos z}{z^8}\,dz$

We know that, $$h(z)=1+\cos z$$ is a holomorphic function and that the point $$z=0$$ is contained inside the closed contour $$C:~| z - 1 | = 2$$. Therefore by the Cauchy's integral formula we get,

$\oint_C g(z)\,dz=\oint_C\frac{1+\cos z}{z^8}\,dz=\frac{2\pi i}{7!}h^{7}(0)$

Now can you do the first part yourself? Kind Regards,
Sudharaka.

#### chisigma

##### Well-known member
Determine the location and nature of singularities in the finite z plane of the following functions:
(a) f(z) = (
- 1) sin(z)/[z(z+1)(z+2)(z-3)]

Using Cauchy's intergral formulae, referring to the above function, evaluate
with C : | z + j | = 4 , traversed positively (CCW)...
The function...

$\displaystyle f(z)= \frac {(z^{2}-1)\ \sin z}{z\ (z+1)\ (z+2)\ (z-3)}\$ (1)

... two poles in $z=-2$ and $z=3$ and two so called 'removable singularities' [a concept that in my opinion produces only confusion and that should be removed from the text books...] in $z=0$ and $z=-1$. The residues of f(z) are...

$R_{-2}= \lim_{z \rightarrow -2} (z+2)\ f(z)= \frac {3}{10}\ \sin 2$

$R_{+3}=\lim_{z \rightarrow +3} (z-3)\ f(z)= \frac{2}{15}\ \sin 3$

... and both the poles are inside C so that the required integral is...

$\displaystyle \int_{C} f(z)\ dz = 2\ \pi\ i\ (R_{-2}+ R_{+3})$ (2)

Kind regards

$\chi$ $\sigma$

Last edited: