 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/7/d/solution

w = z - 1

\frac{2 z}{z ^{2} + 1} = \frac{1}{z - i} + \frac{1}{z + i} = \frac{1}{1 - i + w} + \frac{1}{1 + i + w} .

f (z) = \sum_{n = 0}^{\infty} (- 1)^{n} [(1 - i)^{- n - 1} + (1 + i)^{- n - 1}] w^{n} .

The two bracketed terms are complex conjugates. Using

1 \pm i = 2 e^{\pm iπ /4}

produces the required real coefficients:

f (z) = n = 0 \sum \infty (- 1)^{n} 2^{(1 - n) /2} cos (\frac{( n + 1 ) π}{4}) (z - 1)^{n} .

The initial terms are

1 - \frac{1}{2} w^{2} + \frac{1}{2} w^{3} - \frac{1}{4} w^{4} + \dots

z = \pm i

, both at distance

2

from the center. Hence the Taylor series has

R = 2, ∣ z - 1∣ < 2 .

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