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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/7/c/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 7 / c / Solution by

Codex  0 2026-10-18

Recognize that

z^{2} = (x^{2} - y^{2}) + 2 i x y

and

e^{- z} = e^{- x} (cos y - i sin y)

. Their product has exactly the specified real part. Its imaginary part is

v_{0} = e^{- x} [2 x y cos y - (x^{2} - y^{2}) sin y] .

One can also recover it directly from the Cauchy-Riemann equations. Differentiating the specified real part gives

u_{x} = e^{- x} [(2 x - x^{2} + y^{2}) cos y + 2 y (1 - x) sin y] .

Integration of

v_{y} = u_{x}

gives

v = v_{0} + h (x)

; the other equation

v_{x} = - u_{y}

h^{'} (x) = 0

. Thus the most general answer on a connected domain is

f (z) = z^{2} e^{- z} + i C, C \in R .

Indeed the difference of any two analytic functions with the same real part has zero real part; its two Cauchy-Riemann equations force its imaginary part to be constant. This proves that no further analytic freedom is missing.

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