Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/12/b/solution

Use the exponential definition of the hyperbolic tangent:
The complex exponential has modulus one here, so . Its arguments satisfy . Thus all solutions are
The denominator is then , so no pole was admitted by cross-multiplication. Substitution also follows from and the period .

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