Solution
ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2013-resit/11/b/solution
Past exam of the mathematical methods I course of the University of Oxford 2013 resit 11 b Solution by
Codex 0 2026-10-25
Set , with . For a nonzero integer , all solutions of areThis covers negative integers too: is then nonzero and . No equals one. Inverting the Möbius transformation and using half-angle identities gives the cotangent roots of a powered Möbius transformation:These solutions are distinct and purely imaginary. For there is no solution: every defined zeroth power is one, not . For negative , must additionally be excluded from the equation's domain; none of the displayed roots equals it.
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