Question

Locate and name the singularities of the following functions in the finite z-plane:
i) ln(z+3i)/z²
ii) (z²-2z)/(z²+2z+2)

Answer 1

Locate and name the singularities of the following functions in the finite z-plane:

1. $f_{1}(z)=\frac{\ln(z+3i)}{z^{2}}$

2. $f_{2}(z)=\frac{z^{2}-2z}{z^{2}+2z+2}$

Answer:

1. the function $f_{1}(z)$ has 2 singularities at points: $-3i$ and 0.

2. the function $f_{2}(z)$ has 2 singularities at points: $-1+i$ and $-1-i$.

1 $f_{1}(z)$

A function $\ln(z)$ has a singularity at $z=0$. This type of singularity points is also called a branch point. It means that as we travel from the point to itself looping around the branch point, we will eventually change the function value.

Now we know that $\ln(z+3i)$ has a singularity point at $z=-3i$. In the vicinity of this point function $\frac{1}{z^{2}}$ is analytical. So the multiplication of those two will still has a singularity.

A function $\frac{1}{z^{2}}$ has a singularity at $z=0$, namely a pole of order 2. In the vicinity of this point function $\ln(z+3i)$ is analytical. As in the previous case, we conclude that $f_{1}(z)$ has one more singularity.

As the function $f_{1}(z)$ is analytic everywhere except those two points, we now conclude that no more singularities exist.

2 $f_{2}(z)$

Let’s start with transforming the function:

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

$=\frac{z(z-2)}{(z+1)^{2}-i^{2}}=\frac{z(z-2)}{(z+1-i)(z+1+i)}$

$=z\frac{1+3i}{2(z+1-i)}+z\frac{1-3i}{2(z+1+i)}$

Now we see that at points $-1+i$ and $-1-i$ function $f_{2}(z)$ has poles of order 1.

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