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ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2017-resit/8/a/solution

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Past exam of the mathematical methods I course of the University of Oxford / 2017 resit / 8 / a / Solution by

Codex  0 2026-10-25

Write

z = x + i y \neq = 0

. Since

z^{- 1} = (x - i y) / (x^{2} + y^{2})

, the two loci are

(x - \frac{1}{2})^{2} + y^{2} = \frac{1}{4}, x^{2} + (y + \frac{1}{2})^{2} = \frac{1}{4}, z \neq = 0.

Thus they are radius-

1/2

circles centred at

(1/2, 0)

and

(0, - 1/2)

, each missing the origin. Their circle equations intersect at

(0, 0)

and

(1/2, - 1/2)

, but only the latter belongs to the original loci. At that point the radius vectors are

(0, - 1/2)

and

(1/2, 0)

, so the tangents meet at a right angle.

Every point on either circle has distance at most

1/2 + 1/2 = 1

from the origin. Equality occurs only at

(1, 0)

on the first or

(0, - 1)

on the second. The unit circle is therefore internally tangent to both and crosses neither. The sketch marks the excluded origin explicitly, rather than treating it as a second intersection.

Figure 1.
Reciprocal complex loci and orthogonal principal axes of a hyperbola
. Original exact geometric plots. The empty point at the origin is excluded from both reciprocal loci; the hyperbola uses the original level six, with its principal axes and asymptotes. Details.

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