Question #85328

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

2. z^2-2z/(z^2+2z+2)

Expert's answer

Answer to Question #85328 – Math – Complex Analysis

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

Question

1. ln(z+3i)/z2\ln(z+3i)/z^2

Solution

1. f(z)=ln(z+3i)z2f(z) = \frac{\ln(z+3i)}{z^2}

This function has two singularities: one at z=0z=0 of order 2 and other at z+3i=0z+3i=0 or z=3iz=-3i.

Z=0Z=0 is the pole of order 2.

Function of 1/Z21/Z^2 has the singularity at z=0z=0, pole of order 2.

Function of ln(z+3i)\ln(z+3i) has a singularity point at z=3iz=-3i, singularity point is branch point.

Question

2. z22z/(z2+2z+2)z^2 - 2z/(z^2 + 2z + 2)

Solution

2. f(z)=z22zz2+2z+2f(z) = \frac{z^2 - 2z}{z^2 + 2z + 2}

f(z)=z(z2)(z+1)2+1f(z) = \frac{z(z-2)}{(z+1)^2 + 1}f(z)=z(z2)(z+1)2i2f(z) = \frac{z(z-2)}{(z+1)^2 - i^2}f(z)=z(z2)(z+1i)(z+1+i)f(z) = \frac{z(z-2)}{(z+1-i)(z+1+i)}


In order to find pole, take denominator equal to zero.

Thus, z+1i=0z + 1 - i = 0 and z+1+i=0z + 1 + i = 0

Or, z=1+Iz = -1 + I and z=1iz = -1 - i

Function has a singularity of the pole of order 1 at z=1+Iz = -1 + I and z=1iz = -1 - i.

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