Re: Is Mandelbrot Set symmetrical around the imaginary axis?

Hm... actually for Julia sets its quite more simple. The Zn(zo)
expansion can be easilly check to be a power series of zo with only
even powers. Since for any complex number z, z^2 = (-z)^2, the the
complete expansion happens to take the same value for zo and -zo. So
does the modulus of the expansion too (|Zn|), and thus the complete
Julia set is "symmetrical" under change of sign in the dynamica plane
(zo).

Iñigo Quilez

.

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