 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/7/b/solution

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

v_{y} = u_{x}

integrates with respect to

y

v = e^{x} (x sin y + y cos y) + h (x) .

Now

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

v_{x} = - u_{y}

h^{'} (x) = 0

. Hence

f = e^{x} {x cos y - y sin y + i (x sin y + y cos y)} + i C = z e^{z} + i C .

The normalization at zero gives

C = 0

, so the required entire function is

f (z) = z e^{z} .

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