Lemniscate of Bernoulli as Locus in Complex Plane

Theorem

The locus of $z$ on the complex plane such that:

$\cmod {z - a} \cmod {z + a} = a^2$

is a lemniscate of Bernoulli.

Proof

\(\ds \cmod {z - a} \cmod {z + a}\)

\(=\)

\(\ds a^2\)

\(\ds \leadsto \ \ \)

\(\ds \paren {\paren {x - a}^2 + y^2} \paren {\paren {x + a}^2 + y^2}\)

\(=\)

\(\ds a^4\)

Definition of Complex Modulus

\(\ds \leadsto \ \ \)

\(\ds \paren {x^2 - 2 a x + a^2 + y^2} \paren {x^2 + 2 a x + a^2 + y^2}\)

\(=\)

\(\ds a^4\)

\(\ds \leadsto \ \ \)

\(\ds \paren {\paren {x^2 + a^2 + y^2} - 2 a x} \paren {\paren {x^2 + a^2 + y^2} + 2 a x}\)

\(=\)

\(\ds a^4\)

\(\ds \leadsto \ \ \)

\(\ds \paren {x^2 + a^2 + y^2}^2 - 4 a^2 x^2\)

\(=\)

\(\ds a^4\)

Difference of Two Squares

\(\ds \leadsto \ \ \)

\(\ds \paren {x^2 + y^2}^2 + a^4 + 2 \paren {x^2 + y^2} a^2 - 4 a^2 x^2\)

\(=\)

\(\ds a^4\)

\(\ds \leadsto \ \ \)

\(\ds \paren {x^2 + y^2}^2 + 2 a^2 \paren {x^2 - 2 x^2 + y^2}\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \paren {x^2 + y^2}^2 + 2 a^2 \paren {y^2 - x^2}\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \paren {x^2 + y^2}^2\)

\(=\)

\(\ds 2 a^2 \paren {x^2 - y^2}\)

which is the equation defining a lemniscate of Bernoulli with major semiaxis $a \sqrt 2$.

$\blacksquare$

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $141$