Singularities of Complex Functions

[timeforchg1]

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]

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]

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 textbooks...] 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

$\chi$ $\sigma$