 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s1/2/d/solution

ρ > 0

, write

z = ρ e^{i θ}

. Then

w = (ρ + ρ^{- 1}) cos θ + i (ρ - ρ^{- 1}) sin θ .

For

ρ \neq = 1

its image is the ellipse

\frac{( Re w ) ^{2}}{( ρ + ρ ^{- 1} ) ^{2}} + \frac{( Im w ) ^{2}}{( ρ - ρ ^{- 1} ) ^{2}} = 1.

Its horizontal semiaxis is

ρ + ρ^{- 1}

, its vertical semiaxis is

∣ ρ - ρ^{- 1} ∣

, and its foci are

\pm 2

because the squared semiaxes differ by

4

. Reciprocal radii give the same ellipse, traversed in opposite directions as

θ

increases.

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