Question 3.25: For each of the following functions, locate and name the sin...
For each of the following functions, locate and name the singularities in the finite z plane and determine whether they are isolated singularities or not.
(a) f(z)=\frac{z}{\left(z^2+4\right)^2}, (b) f(z)=\sec (1 / z) (c) f(z)=\frac{\ln (z-2)}{\left(z^2+2 z+2\right)^4}, (d) f(z)=\frac{\sin \sqrt{z}}{\sqrt{z}}
(a) f(z)=\frac{z}{\left(z^2+4\right)^2}=\frac{z}{\{(z+2 i)(z-2 i)\}^2}=\frac{z}{(z+2 i)^2(z-2 i)^2}.
Since
z = 2i is a pole of order 2. Similarly, z = -2i is a pole of order 2.
Since we can find \delta such that no singularity other than z=2 i lies inside the circle |z-2 i|=\delta (e.g., choose \delta=1 ), it follows that z=2 i is an isolated singularity. Similarly, z=-2 i is an isolated singularity.
Since sec(1/z) = 1/cos(1/z), the singularities occur where cos(1/z) = 0, i.e., 1/z =(2n+1)Ï€/2 or z =2/(2n+1)Ï€, where n =0, ±1, ±2, ±3, … . Also, since f(z) is not defined at z = 0, it follows that z = 0 is also a singularity.
Now, by L’Hospital’s rule,
Thus the singularities z = 2/(2n + 1)/Ï€, n = 0, ±1, ±2, … are poles of order one, i.e., simple poles. Note that these poles are located on the real axis at z =±2/Ï€, ±2/3Ï€, ±2/5Ï€, … and that there are infinitely many in a finite interval which includes 0 (see Fig. 3-9).
Since we can surround each of these by a circle of radius δ, which contains no other singularity, it follows that they are isolated singularities. It should be noted that the δ required is smaller the closer the singularity is to the origin.
Since we cannot find any positive integer n such that \lim _{z \rightarrow 0}(z-0)^n f(z)=A \neq 0, it follows that z=0 is an essential singularity. Also, since every circle of radius \delta with center at z=0 contains singular points other than z=0, no matter how small we take \delta, we see that z=0 is a non-isolated singularity.
(c) The point z = 2 is a branch point and is a non-isolated singularity. Also, since z² + 2z + 2 = 0 where z =-1 ± i, it follows that z²+2z + 2 = (z + 1 + i)(z + 1 – i) and that z = -1 ± i are poles of order 4 which are isolated singularities
(d) At first sight, it appears as if z=0 is a branch point. To test this, let z=r e^{i \theta}=r e^{i(\theta+2 \pi)} where 0 \leq \theta<2 \pi.
If z=r e^{i \theta}, we have
f(z)=\frac{\sin \left(\sqrt{r} e^{i \theta / 2}\right)}{\sqrt{r} e^{i \theta / 2}}If z=r e^{i(\theta+2 \pi)}, we have
f(z)=\frac{\sin \left(\sqrt{r} e^{i \theta / 2} e^{\pi i}\right)}{\sqrt{r} e^{i \theta / 2} e^{\pi i}}=\frac{\sin \left(-\sqrt{r} e^{i \theta /2}\right)}{-\sqrt{r} e^{i \theta / 2}}=\frac{\sin \left(\sqrt{r} e^{i \theta / 2}\right)}{\sqrt{r} e^{i \theta / 2}}Thus, there is actually only one branch to the function, and so z=0 cannot be a branch point. Since \lim _{z \rightarrow 0} \sin \sqrt{z} / \sqrt{z}=1, it follows in fact that $z=0$ is a removable singularity.
