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Solutions for assignment 6

Solutions for assignment 6 - Solutions for assignment 6 1...

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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)

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Solutions for assignment 6 - Solutions for assignment 6 1...

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