a) Find the principal part at of the function
Answer: The part shown is the principal part.
b) Find all the singularities of in the disk . Determine the nature of each singularity (isolated, removable, pole of what order, essential).
Answer: Zero is a pole of order . The function can be written in the form , where Integral multiples of are simple zeroes (of order ) of the function , of which , , and are in the disk . The function does not vanish at and . Hence, and are simple poles of .
c) Find the residue at each isolated singularity in .
Answer: Using the Laurent serries above, we see that . At the simple pole , the residues is
vanishes to order at integer multiples of
The numerator has value at . Thus,
has a simple pole at , .
The only multiple of enclosed by is .
Using the fact that is a simple pole, we get
is an entire function, so its integral, over any closed
contour, is zero (by Cauchy-Goursat's Theorem).
Using the parametrization
, we get
Let . Then
The integral gets converted to the contour integral over the unit circle .
Answer: See the first Example in Section 60 of the text on page 205.
be the Taylor series of
centered at . Taylor's Theorem states, in particular, that
for . We get, for ,
b) Find the Taylor series of the function around the point .
Use the partial fraction decomposition
a) The limit exists and is equal to .
Answer: False, is not analytic at . The above limit is the derivative limit, which does not exists. A direct argument, that the limit doesn't exists, consists of letting approach along the and axis:
There is a function , analytic in the disk
Answer: False. Such a function would be a non-constant analytic function, whose absolute value achieves its maximum at the interior point of . This contradicts the Maximum Modulus Principle.
c) If has an isolated singularity at and , then is a removeable singularity.
Answer: False. Take and as a counter example.
The curve is parametrized by
has a positive imaginary part. Consequently,
has a negative real part equal to
Note the equality