Solution tutorial exercise 3 (part 2)

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.

http://math.fullerton.edu/mathews/c2003/SingularityZeroPoleMod.html

Solution 3 (b).


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).


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

http://math.fullerton.edu/mathews/c2003/LaurentSeriesMod.html





Then apply Definition 7.5 to conclude that has an essential


We are done.

Solution 3 (d).


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).




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).


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).




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


We are done.

Solution 3 (h).


Answer. has a pole of order 2 at the

origin.


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.



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Solution tutorial exercise 3 (part 2)