 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/1/solution

z^{2} + 1 = (z - i) (z + i)

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

(z - i) (z + i)

and comparison of polynomial coefficients give

A + B = 1

and

i (A - B) = 1

. Hence

\frac{z + 1}{z ^{2} + 1} = \frac{1 - i}{2 ( z - i )} + \frac{1 + i}{2 ( z + i )}, z \neq = \pm i .

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