 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-3/2f/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

4 a^{3} + 27 b^{2} = 0

and

(a, b) \neq = (0, 0)

, there is a nonzero real number

r

such that

a = - 3 r^{2}, b = 2 r^{3} .

Then

(0, 0, r)

is singular and

z^{3} + a z + b = (z - r)^{2} (z + 2 r) .

For

r > 0

, the singular point is locally isolated because all terms in

x^{2} + y^{2} + (z - r)^{2} (z + 2 r) = 0

are nonnegative nearby. For

r < 0

, the surface is locally a double cone with vertex at that point. Neither neighbourhood is homeomorphic to an open disc, so

S_{a, b} is not a smooth surface when 4 a^{3} + 27 b^{2} = 0.

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