## Property of R. Bryant's parametrizationEdit

If *z* is replaced by the negative reciprocal of its complex conjugate, then the functions *g _{1}*,

*g*, and

_{2}*g*of

_{3}*z*are left unchanged.

### ProofEdit

Let *g _{1}′* be obtained from

*g*by substituting

_{1}*z*with Then we obtain

Multiply both numerator and denominator by

Multiply both numerator and denominator by -1,

It is generally true for any complex number *z* and any integral power *n* that

therefore

therefore since, for any complex number *z*,

Let *g _{2}′* be obtained from

*g*by substituting

_{2}*z*with Then we obtain

therefore since, for any complex number *z*,

Let *g _{3}′* be obtained from

*g*by substituting

_{3}*z*with Then we obtain

therefore Q.E.D.

## Symmetry of the Boy's surfaceEdit

Boy's surface has 3-fold symmetry. This means that it has an axis of discrete rotational symmetry: any 120° turn about this axis will leave the surface looking exactly the same. The Boy's surface can be cut into three mutually congruent pieces.

### ProofEdit

Two complex-algebraic identities will be used in this proof: let *U* and *V* be complex numbers, then

Given a point *P(z)* on the Boy's surface with complex parameter *z* inside the unit disk in the complex plane, we will show that rotating the parameter *z* 120° about the origin of the complex plane is equivalent to rotating the Boy's surface 120° about the *Z*-axis (still using R. Bryant's parametric equations given above).

Let

be the rotation of parameter *z*. Then the "raw" (unscaled) coordinates *g _{1}*,

*g*, and

_{2}*g*will be converted, respectively, to

_{3}*g′*,

_{1}*g′*, and

_{2}*g′*.

_{3}Substitute *z′* for *z* in *g _{3}(z)*, resulting in

Since it follows that

therefore This means that the axis of rotational symmetry will be parallel to the *Z*-axis.

Plug in *z′* for *z* in *g _{1}(z)*, resulting in

Noticing that

Then, letting in the denominator yields

Now, applying the complex-algebraic identity, and letting

we get

Both and are distributive with respect to addition, and

due to Euler's formula, so that

Applying the complex-algebraic identities again, and simplifying to -1/2 and to produces

Simplify constants,

therefore

Applying the complex-algebraic identity to the original *g _{1}* yields

Plug in *z′* for *z* in *g _{2}(z)*, resulting in

Simplify the exponents,

Now apply the complex-algebraic identity to *g′ _{2}*, obtaining

Distribute the with respect to addition, and simplify constants,

Apply the complex-algebraic identities again,

Simplify constants,

then distribute with respect to addition,

Applying the complex-algebraic identity to the original *g _{2}* yields

The raw coordinates of the pre-rotated point are

and the raw coordinates of the post-rotated point are

Comparing these four coordinates we can verify that

In matrix form, this can be expressed as

Therefore rotating *z* by 120° to *z′* on the complex plane is equivalent to rotating *P(z)* by -120° about the *Z*-axis to *P(z′)*. This means that the Boy's surface has 3-fold symmetry, *quod erat demonstrandum*.