Question 2: The limit of the ratios of the terms in \sum a_j (z-z_0)^j is L |z-z_0|, because of the given information that lim_j |a_{j+1} / a_j| = L. Thus, by the ratio test, the series converges when |z-z_0| < 1/L and diverges when |z-z_0| > 1/L. Thus the radius of convergence of the series is 1/L.

Question 3 (b). L = 2, so the radius of convergence is 1/2, and the circle of convergence is |z - 1| = 1/2.

(d) L = 1/3, so the radius of convergence is 3, and the circle of convergence is |z - i| = 3.

Section 5.5

Question 3: Although it is not mentioned explicitly, it is assumed that we are computing the Laurent series aroud zero.

The easiest way is to break the function up into partial fractions:

z/(z+1)(z-2) = (1/3)/(z+1) + (2/3)/(z-2);You can either work out the constants 1/3, 2/3 by the usual method taught to you in lower-division, or you can compute the residues of z/(z+1)(z-2) at -1 and 2 respectively.

(a) When |z|<1 we have

1/(z+1) = 1/(1 - (-z)) = 1 - z + z^2 - z^3 + ...and

1/(z-2) = (-1/2) 1/(1 - z/2) = (-1/2) (1 + z/2 + z^2 / 4 + ...)so that f is equal to

f(z) = 1/3 (1 - z + z^2 - z^3 + ...) + (2/3) (-1/2) (1 + z/2 + z^2 / 4 + ...)

= - 1/2 z + 1/4 z^2 - 3/8 z^3 + ...You can also use sigma notation to obtain the answer in the back of the book.

(b) When 1<|z|<2 we still have

1/(z-2) = (-1/2) (1 + z/2 + z^2 / 4 + ...)but the formula for 1/(z+1) no longer works (because the geometric series formula does not apply), and we need a different Laurent series:

1/(z+1) = (1/z) 1/(1 - (-1/z)) = 1/z(1 - 1/z + 1/z^2 - 1/z^3 + ...)Thus f(z) becomes

f(z) = -1/3 - 1/6 z - 1/12 z^2 - ... + 1/(3z) - 1/(3z^2) + 1/(3z^3) + ...)

(c) When |z|>2 the formula

1/(z+1) = 1/z (1 - (-1/z)) = 1/z(1 - 1/z + 1/z^2 - 1/z^3 + ...)still works (since -1/z still has magnitude less than 1), but the formula for 1/z-2 has to be replaced, since z/2 no longer has magnitude less than 1:

1/(z-2) = (1/z) 1/(1 - 2/z) = (1/z)(1 + 2/z + 4/z^2 + 8/z^3 + ...)iand so we get

f(z) = 1/3(1/z(1 - 1/z + 1/z^2 - 1/z^3 + ...)) + (2/3) ((1/z)(1 + 2/z + 4/z^2 + 8/z^3 + ...))

= 1/z - 1/z^2 + 3/z^3 - 5/z^4 + ...

Question 4: Although it is not mentioned explicitly, it is assumed that the Laurent expansion requested is around zero.

By replacing z by 2z in the formula for sin(z) we obtain

sin(2z) = 2z - (2z)^3/3! + (2z)^5/5! - ...and so when |z| > 0 we get

sin(2z)/z^3 = 2/z^2 - 2^3/3! + 2^5 z^2/5! - ...which is in Laurent series form.

Question 7(b): This function has a simple pole at 1, and is analytic on the annulus $0 < |z| < 2pi$, so we know that

1/(e^z - 1) = a_-1 / z + a_0 + a_1 z + a_2 z^2 + ...for all z in this annulus. Multiplying both sides by e^z - 1 we get

1 = (a_-1 / z + a_0 + a_1 z + a_2 z^2 + ...) (z + z^2/2! + z^3/3! + ...)and by comparing co-efficients we get

a_-1 = 1

a_-1 / 2! + a_0 = 0

a_-1 / 3! + a_0/2! + a_1 = 0

...so that

a_-1 = 1, a_0 = -1/2, a_1 = 1/12, etc.Thus the first few terms of the Laurent series are

1/z - 1/2 + z/12 + ...

Question 9: The series can be broken up into two parts, the positive power part

1 + z/2 + z^2/4 + ...and the negative power part

1/2z + 1/4z^2 + ...The positive power part is a geometric series that converges when |z| < 2 and diverges when |z| > 2. The negative power part is also a geometric series (with the constant term removed), and it converges when |1/2z| < 1, i.e when |z| > 1/2, and diverges when |z| < 1/2. Thus the series as a whole converges on the annulus 1/2 < |z| < 2, and diverges outside this annulus (since a convergent series plus a divergent series is always divergent). On the boundaries |z| = 1/2 and |z| = 2 of the annulus the question is more complicated; it turns out that the series does not converge on the boundary because the terms do not go to zero.