Section 5.1

Question 1(b): [I use \sum to denote the Sigma summation sign] This is 3 \sum z^k with z = 1/(1+i), so by the geometric series formula and the fact that |z| < 1, we obtain 3/(1 - z) = 3/(1 - 1/(1+i)) = 3 - 3i.

(c) Similarly, the expression is \sum z^k with z = -2/3, so we obtain 1 / (1 - z) = 3/5.

Question 3: If z_n converges to z (say), then z_{n-1} also converges to z. So lim (z_n - z_{n-1}) = lim z_n - lim z_{n-1} = z - z = 0.

Question 7: (a-e) use the Ratio test and find the limit of |z_{n+1}|/|z_n|, using the same techniques you would use in real-variable calculus. For (f), note that the summands do not go to zero (in fact | i^k - 1/k^2 | converges to 1, not zero), so the series has to diverge.

Section 5.2

Question 1(a): If f(z) = exp(-z), then the j^th derivative of f(z) is exp(-z) if j is even and -exp(-z) if j is odd. Thus f^(j)(0) = (-1)^j, and the result follows by Taylor's formula (at z_0 = 0).

(e): If f(z) = sinh(z), then the j^th derivative of f(z) is sinh(z) if j is even and cosh(-z) if j is odd. Thus f^(2j)(0) = 0 and f^{2j+1}(0) = 1, and the result follows by Taylor's formula (at z_0 = 0).

Question 2: You could do this by the ratio test, but there is a much simpler way. Both functions are analytic on the entire complex plane, so the radius of convergence of both power series is infinite. (If, for example, exp(-z) was only analytic on a disk of radius 4 around the origin, then you would only know that the power series converged for |z| < 4, and in all probability it would not converge outside of this disk.)

Question 8(a) The Taylor series for both sides is equal to

-z + z^3/3! - z^5/5! + ...so the two sides must be equal.

(c) The Taylor series for both sides is equal to

1 - iz - z^2/2! + iz^3/3! + z^4/4! - iz^5/5! - ...so the two sides must be equal.

Question 11(a) Applying the formula (13) mentioned in Theorem 6 to the two Maclaurin series

exp(z) = 1 + z + z^2/2! + z^3/3! + ...

cos(z) = 1 + 0z - z^2/2! + 0z^3 + ...we see that the first few co-efficients are

c_0 = a_0 b_0 = 1 x 1 = 1

c_1 = a_0 b_1 + a_1 b_0 = 1 x 0 + 1 x 1 = 1

c_2 = a_0 b_2 + a_1 b_1 + a_2 b_0 = 1 x -1/2 + 1 x 0 + 1/2 x -1 = 0

c_3 = 1 x 0 + 1 x -1/2 + 0 x 1/2 + 1 x 1/3! = -1/3so that the first three non-zero terms are

exp(z)cos(z) = 1 + z - z^3/3 + ...Of course, it is much more intuitive simply to multiply out the two Taylor series together as if they were finite polynomials. Theorem 6 merely gives a theoretical guarantee that you will get the right answer from doing that.

(b) Doing the same thing as (a) to the Maclaurin series

exp(z) = 1 + z + z^2/2! + z^3/3! + ...

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

exp(z)/(1-z) = 1 + 2z + 5/2 z^2 + ...

Question 14: Since f is analytic in D, it has a Taylor series expansion

f(z) = f(z_0) + f'(z_0) (z-z_0) + ...which is valid in D. However by hypothesis every single coefficient on the right-hand side is zero. Thus f(z) is zero everywhere in D.