Section 5.6
Question 1: (a) The function is analytic everywhere except at 0 and 1. The denominator has a double zero at
0 and a simple zero at 1, and the numerator has a simple zero at 1 and no zero at 0, so the quotient has a
double pole at 0 and a removable singularity at 1.
(c) 4z is always analytic, so it does not affect any singularities. the denominator has simple zeroes at +1 and
1, while the numerator is nonzero at those points, so the function as a whole has a simple pole at +1 and 1.
(d) e^z  1 has zeroes at all multiplies of 2pi i, and they are all simple (since the derivative of e^z1 is nonzero
at these points). So the function has simple poles at all multiples of 2 pi i.
(g) The only place where the function is not analytic is at 0. The easiest way to do this one is by finding the
Laurent series. Rewriting the function as (sin(3z)  3z) / z^2, we see that the numerator is  (3z)^3/3! +
(3z)^5/5!  .
.., so the Laurent series of the function as a whole starts with a linear (z^1) term. In particular,
there are no negative power terms, so the function has a removable singularity, and because the first term is
z^1, the function has a simple zero once the singularity is removed.
Question 2: 2 cos z  2 + z^2 has a zero of order 4 at 0, as one can see by a Taylor series expansion (or by
differentiation). Thus, the denominator of f has a zero of order 8, and f itself has a pole of order 8 at 0.
Question 3: See examples at back of book.
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 Spring '08
 Grossman
 Laurent, removable singularity

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