The document provides solutions to 8 exercises analyzing the types of singularities (removable, pole, essential) of various functions. For example, it is shown that the function has a simple pole at the origin by finding where the denominator is zero and applying Corollary 7.5, while the function has an essential singularity at the origin by expressing it as a Laurent series and applying Definition 7.5. Other solutions determine singularities as removable, simple poles at specific points, or a pole of order 2.

1. Solution 3 (a).
See text and/or instructor's solution manual.
Answer. has a simple pole at the origin.
Solution. Use a convenient method to determine where the denominator is zero.
For example, use the series and get
Then
Now apply Corollary 7.5 to conclude that has a simple pole at the
origin.
You do not need to find any other singularity for this exercise.
We are done.

2. Solution 3 (b).
See text and/or instructor's solution manual.
Answer. has an essential singularity at the origin.
Solution. Use the fact that , and express this function as
a Laurent series.
Use the substitution and obtain
Then apply Definition 7.5 to conclude that has an essential
singularity at the origin.
We are done.
Solution 3 (c).
See text and/or instructor's solution manual.
Answer. has an essential singularity at the origin.
Solution. Use the fact that , and express this function as a Laurent
series.
Use the substitution and get

3. Then apply Definition 7.5 to conclude that has an essential
singularity at the origin.
We are done.
Solution 3 (d).
See text and/or instructor's solution manual.
Answer. has simple poles at the points where n is an
integer.
Solution. We know that has simple zeros at
for .
Now apply Corollary 7.5 to conclude that has simple poles
at for .
We are done.
Solution 3 (e).
See text and/or instructor's solution manual.

4. Answer. has removable a singularity at the origin, and a simple
pole at -1.
Solution. Write
We know that has a removable singularity at ,
which implies that that has a removable singularity at
Now apply Corollary 7.5 to conclude that has a simple
pole at -1.
Therefore, has removable a singularity at
the origin, and a simple pole at -1.
We are done.
Solution 3 (f).
See text and/or instructor's solution manual.
Answer. has simple poles at the points where , and
a removable singularity at the origin.
Solution. We know that has simple zeros at for ,
and that has a removable singularity at the origin.
Now apply Corollary 7.5 to conclude that has a simple poles at
for ,
also has a removable singularity at the origin.
We are done.
Solution 3 (g).

5. See text and/or instructor's solution manual.
Answer. has a removable singularity at the origin if we
define .
Solution. Consider . Use the known series
and write
.
Substitute this series in the numerator and obtain
Then apply Definition 7.5 to conclude that has a removable
singularity at the origin.
We are done.
Solution 3 (h).
See text and/or instructor's solution manual.
Answer. has a pole of order 2 at the
origin.

6. Solution. Use the fact that , and
and express this function as a Laurent series.
Then apply Definition 7.5 to conclude
that has a pole of order 2 at the
origin.
We are done.