Solutions for assignment 8

# Solutions for assignment 8 - Solutions for assignment 8 1...

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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 non-zero 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^z-1 is non-zero 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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Solutions for assignment 8 - Solutions for assignment 8 1...

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