If z is replaced by the negative reciprocal of its complex conjugate,
then the functions g1, g2, and g3 of z are left unchanged.
Let g1′ be obtained from g1 by substituting z with
Then we obtain
![{\displaystyle g_{1}'=-{3 \over 2}\mathrm {Im} \left({-{1 \over z^{\star }}\left(1-{1 \over z^{\star 4}}\right) \over {1 \over z^{\star 6}}-{\sqrt {5}}{1 \over z^{\star 3}}-1}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00e38ec1fbf74ffd568f08dbcdca10883125ea03)
Multiply both numerator and denominator by
![{\displaystyle g_{1}'=-{3 \over 2}\mathrm {Im} \left({-z^{\star }(z^{\star 4}-1) \over 1-{\sqrt {5}}z^{\star 3}-z^{\star 6}}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af8936de6cdb94d982fbb7de4ce450fb973fa4b1)
Multiply both numerator and denominator by -1,
![{\displaystyle g_{1}'=-{3 \over 2}\mathrm {Im} \left({z^{\star }(z^{\star 4}-1) \over z^{\star 6}+{\sqrt {5}}z^{\star 3}-1}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/619c841d4b5681bbb6b32c340e15cbb20db54ce1)
It is generally true for any complex number z and any integral power n that
![{\displaystyle (z^{\star })^{n}=(z^{n})^{\star },}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbec06b7a7f6d444f420fff594cb1ebccd515e46)
therefore
![{\displaystyle g_{1}'=-{3 \over 2}\mathrm {Im} \left({z^{\star }(z^{4}-1)^{\star } \over (z^{6}+{\sqrt {5}}z^{3}-1)^{\star }}\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edd3a4f3253845f9bc3ab48257874db6b1d8b245)
![{\displaystyle g_{1}'=-{3 \over 2}\mathrm {Im} \left(-\left({z(1-z^{4}) \over z^{6}+{\sqrt {5}}z^{3}-1}\right)^{\star }\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1467af869ae3f028c3c38ff079d8342a53a2a796)
therefore
since, for any complex number z,
![{\displaystyle \mathrm {Im} (-z^{\star })=\mathrm {Im} (z).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68dfd7969d7b1a566647807953169f86e2dbe544)
Let g2′ be obtained from g2 by substituting z with
Then we obtain
![{\displaystyle =-{3 \over 2}\mathrm {Re} \left({z^{\star }(z^{\star 4}+1) \over z^{\star 6}+{\sqrt {5}}z^{\star 3}-1}\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d14db85c77490c5bb153b7a17879284c566152b7)
![{\displaystyle =-{3 \over 2}\mathrm {Re} \left({z^{\star }(z^{4\star }+1) \over z^{6\star }+{\sqrt {5}}z^{3\star }-1}\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/400cb8de8ecde890023d95c67c7f86b1a97d2254)
![{\displaystyle =-{3 \over 2}\mathrm {Re} \left({z^{\star }(z^{4}+1)^{\star } \over (z^{6}+{\sqrt {5}}z^{3}-1)^{\star }}\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0fe0bcf7d7b606e2ac786866492e034c3bd94d0b)
![{\displaystyle =-{3 \over 2}\mathrm {Re} \left(\left({z(z^{4}+1) \over z^{6}+{\sqrt {5}}z^{3}-1}\right)^{\star }\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94f58913e031463d3fbe8b9fbe0a2fefaf383039)
therefore
since, for any complex number z,
![{\displaystyle \mathrm {Re} (z^{\star })=\mathrm {Re} (z).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a05981d07029f18e1f4f4eb2a41ecf9cb54ed6da)
Let g3′ be obtained from g3 by substituting z with
Then we obtain
![{\displaystyle =\mathrm {Im} \left({z^{\star 6}+1 \over 1-{\sqrt {5}}z^{\star 3}-z^{\star 6}}\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57bb834f9c57f4d0f4be8f5a5d0f6ff7c0c8ed85)
![{\displaystyle =\mathrm {Im} \left({z^{6\star }+1 \over 1-{\sqrt {5}}z^{3\star }-z^{6\star }}\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/091f77870490087051a0e9534468eefa9734d928)
![{\displaystyle =\mathrm {Im} \left(-{(z^{6}+1)^{\star } \over (z^{6}+{\sqrt {5}}z^{3}-1)^{\star }}\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13df1971d6321a1858ecb2d9ecbc0a2beefd11fc)
![{\displaystyle =\mathrm {Im} \left(-\left({z^{6}+1 \over z^{6}+{\sqrt {5}}z^{3}-1}\right)^{\star }\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67d9a08593c8dd3ebbee42afba4540dd1fe9a7b4)
therefore
Q.E.D.
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.
Two complex-algebraic identities will be used in this proof: let U and V be complex numbers, then
![{\displaystyle \mathrm {Re} (UV)=\mathrm {Re} (U)\mathrm {Re} (V)-\mathrm {Im} (U)\mathrm {Im} (V),\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97e2b487422a14b5a95a9659d2cec8b26f9f82c9)
![{\displaystyle \mathrm {Im} (UV)=\mathrm {Re} (U)\mathrm {Im} (V)+\mathrm {Im} (U)\mathrm {Re} (V).\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9519dfaa4b573c194b6f30bef55883161dba503d)
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
![{\displaystyle z'=ze^{i2\pi /3}\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67bf05742158b5efbcf8c271b4421086387a7e3f)
be the rotation of parameter z. Then the "raw" (unscaled) coordinates g1, g2, and g3 will be converted, respectively, to g′1, g′2, and g′3.
Substitute z′ for z in g3(z), resulting in
![{\displaystyle g_{3}'(z')=\mathrm {Im} \left({1+z'^{6} \over z'^{6}+{\sqrt {5}}z'^{3}-1}\right)-{1 \over 2},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c9114a5222dd6e60c8d9beaeba8f8cf1f4f9c49)
![{\displaystyle g_{3}'(z)=\mathrm {Im} \left({1+z^{6}e^{i4\pi } \over z^{6}e^{i4\pi }+{\sqrt {5}}z^{i2\pi }-1}\right)-{1 \over 2}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/444c3270bad0babc68d20487863d534ae5b3bf8d)
Since
it follows that
![{\displaystyle g_{3}'=\mathrm {Im} \left({1+z^{6} \over z^{6}+{\sqrt {5}}z^{3}-1}\right)-{1 \over 2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f32a00cbca6bd45b7d482102bdcb17562bc57c07)
therefore
This means that the axis of rotational symmetry will be parallel to the Z-axis.
Plug in z′ for z in g1(z), resulting in
![{\displaystyle g_{1}'(z)=-{3 \over 2}\mathrm {Im} \left({ze^{i2\pi /3}(1-z^{4}e^{i8\pi /3}) \over z^{6}e^{i4\pi }+{\sqrt {5}}z^{3}e^{i2\pi }-1}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3402a79992a7667c2fc7f42734818a5ac77f44c6)
Noticing that
![{\displaystyle g_{1}'=-{3 \over 2}\mathrm {Im} \left({ze^{i2\pi /3}(1-z^{4}e^{i2\pi /3}) \over z^{6}+{\sqrt {5}}z^{3}-1}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/78178c8d88e1c887e5170c882e579c6bea0f87d6)
Then, letting
in the denominator yields
![{\displaystyle g_{1}'=-{3 \over 2}\mathrm {Im} \left({z(e^{i2\pi /3}-z^{4}e^{-i2\pi /3}) \over z^{6}+{\sqrt {5}}z^{3}-1}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63958a8edbc45efbcdfc50afd48e475aa383e03f)
Now, applying the complex-algebraic identity, and letting
![{\displaystyle z''={z \over z^{6}+{\sqrt {5}}z^{3}-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e507af41e023e6a46c9dc323b1bfe577ec131975)
we get
![{\displaystyle g_{1}'=-{3 \over 2}\left[\mathrm {Im} (z'')\mathrm {Re} (e^{i2\pi /3}-z^{4}e^{-i2\pi /3})+\mathrm {Re} (z'')\mathrm {Im} (e^{i2\pi /3}-z^{4}e^{-i2\pi /3})\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64ddc457f89de8d2e3b4f3c4e8689ad8559447f0)
Both
and
are distributive with respect to addition, and
![{\displaystyle \mathrm {Re} (e^{i\theta })=\cos \theta ,\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91fbb9e5a0e1f3df7240bf6eb90f3ce837755511)
![{\displaystyle \mathrm {Im} (e^{i\theta })=\sin \theta ,\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0050aa3842a3216970fd0dbcf42704ae23d2a62)
due to Euler's formula, so that
![{\displaystyle g_{1}'=-{3 \over 2}\left[\mathrm {Im} (z'')\left(\cos {2\pi \over 3}-\mathrm {Re} (z^{4}e^{-i2\pi /3})\right)+\mathrm {Re} (z'')\left(\sin {2\pi \over 3}-\mathrm {Im} (z^{4}e^{-i2\pi /3})\right)\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a75caf9d08222ef7db660ad6a3af1ffb1b19b87a)
Applying the complex-algebraic identities again, and simplifying
to -1/2 and
to
produces
![{\displaystyle g_{1}'=-{3 \over 2}\left[\mathrm {Im} (z'')\left(-{1 \over 2}-[\mathrm {Re} (z^{4})\mathrm {Re} (e^{-i2\pi /3})-\mathrm {Im} (z^{4})\mathrm {Im} (e^{-i2\pi /3})]\right)+\mathrm {Re} (z'')\left({{\sqrt {3}} \over 2}-[\mathrm {Im} (z^{4})\mathrm {Re} (e^{-i2\pi /3})+\mathrm {Re} (z^{4})\mathrm {Im} (e^{-i2\pi /3})]\right)\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e094e3684fdafbf02d71b5b2f364ae7c33e0a366)
Simplify constants,
![{\displaystyle g_{1}'=-{3 \over 2}\left[\mathrm {Im} (z'')\left(-{1 \over 2}-\left[-{1 \over 2}\mathrm {Re} (z^{4})+{{\sqrt {3}} \over 2}\mathrm {Im} (z^{4})\right]\right)+\mathrm {Re} (z'')\left({{\sqrt {3}} \over 2}-\left[-{1 \over 2}\mathrm {Im} (z^{4})-{{\sqrt {3}} \over 2}\mathrm {Re} (z^{4})\right]\right)\right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/45f95262c48e5b4d8e1fb9295929a3b588ea39d4)
therefore
![{\displaystyle g_{1}'=-{3 \over 2}\left[-{1 \over 2}\mathrm {Im} (z'')+{1 \over 2}\mathrm {Im} (z'')\mathrm {Re} (z^{4})-{{\sqrt {3}} \over 2}\mathrm {Im} (z'')\mathrm {Im} (z^{4})+{{\sqrt {3}} \over 2}\mathrm {Re} (z'')+{1 \over 2}\mathrm {Re} (z'')\mathrm {Im} (z^{4})+{{\sqrt {3}} \over 2}\mathrm {Re} (z'')\mathrm {Re} (z^{4})\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0bb49fb22814d6e3c4c6c64fd35e121a80b0a522)
Applying the complex-algebraic identity to the original g1 yields
![{\displaystyle g_{1}=-{3 \over 2}[\mathrm {Im} (z'')\mathrm {Re} (1-z^{4})+\mathrm {Re} (z'')\mathrm {Im} (1-z^{4})],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80133e0ee3eb4dc621ad7e9434f01f83581072a3)
![{\displaystyle g_{1}=-{3 \over 2}[\mathrm {Im} (z'')(1-\mathrm {Re} (z^{4}))+\mathrm {Re} (z'')(-\mathrm {Im} (z^{4}))],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/73689f32857bd77ded3a212b8be1a94190b0040c)
![{\displaystyle g_{1}=-{3 \over 2}[\mathrm {Im} (z'')-\mathrm {Im} (z'')\mathrm {Re} (z^{4})-\mathrm {Re} (z'')\mathrm {Im} (z^{4})].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7423974f8904a20568735abce4701dd63b536d2f)
Plug in z′ for z in g2(z), resulting in
![{\displaystyle g_{2}'=-{3 \over 2}\mathrm {Re} \left({ze^{i2\pi /3}(1+z^{4}e^{i8\pi /3}) \over z^{6}e^{i4\pi }+{\sqrt {5}}z^{3}e^{i2\pi }-1}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2949f25cfe1537f950d066b2009d6e81d3a95925)
Simplify the exponents,
![{\displaystyle =-{3 \over 2}\mathrm {Re} (z''(e^{i2\pi /3}+z^{4}e^{-i2\pi /3})).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f92308f3a2f030c4c4853281891e0044ce65cbb)
Now apply the complex-algebraic identity to g′2, obtaining
![{\displaystyle g_{2}'=-{3 \over 2}\left[\mathrm {Re} (z'')\mathrm {Re} (e^{i2\pi /3}+z^{4}e^{-i2\pi /3})-\mathrm {Im} (z'')\mathrm {Im} (e^{i2\pi /3}+z^{4}e^{-i2\pi /3})\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/790a2a00eda67d18ad2d98bc277d2a757c36e033)
Distribute the
with respect to addition, and simplify constants,
![{\displaystyle g_{2}'=-{3 \over 2}\left[\mathrm {Re} (z'')\left(-{1 \over 2}+\mathrm {Re} (z^{4}e^{-i2\pi /3})\right)-\mathrm {Im} (z'')\left({{\sqrt {3}} \over 2}+\mathrm {Im} (z^{4}e^{-i2\pi /3})\right)\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/60a5f940997fe5e36745ddea82f9832e2554b442)
Apply the complex-algebraic identities again,
![{\displaystyle g_{2}'=-{3 \over 2}\left[\mathrm {Re} (z'')\left(-{1 \over 2}+\mathrm {Re} (z^{4})\mathrm {Re} (e^{-i2\pi /3})-\mathrm {Im} (z^{4})\mathrm {Im} (e^{-i2\pi \over 3})\right)-\mathrm {Im} (z'')\left({{\sqrt {3}} \over 2}+\mathrm {Im} (z^{4})\mathrm {Re} (e^{-i2\pi /3})+\mathrm {Re} (z^{4})\mathrm {Im} (e^{-i2\pi /3})\right)\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0191d31b2f4d8d75df6b8761cb5a8df325573790)
Simplify constants,
![{\displaystyle g_{2}'=-{3 \over 2}\left[\mathrm {Re} (z'')\left(-{1 \over 2}-{1 \over 2}\mathrm {Re} (z^{4})+{{\sqrt {3}} \over 2}\mathrm {Im} (z^{4})\right)-\mathrm {Im} (z'')\left({{\sqrt {3}} \over 2}-{{\sqrt {3}} \over 2}\mathrm {Re} (z^{4})-{1 \over 2}\mathrm {Im} (z^{4})\right)\right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d610a8f9a991a5eb81e6a533b4d5780468c9020)
then distribute with respect to addition,
![{\displaystyle g_{2}'=-{3 \over 2}\left[-{1 \over 2}\mathrm {Re} (z'')-{1 \over 2}\mathrm {Re} (z'')\mathrm {Re} (z^{4})+{{\sqrt {3}} \over 2}\mathrm {Re} (z'')\mathrm {Im} (z^{4})-{{\sqrt {3}} \over 2}\mathrm {Im} (z'')+{{\sqrt {3}} \over 2}\mathrm {Im} (z'')\mathrm {Re} (z^{4})+{1 \over 2}\mathrm {Im} (z'')\mathrm {Im} (z^{4})\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09aaa064f48fafb2a654719b612c32c7efbc703e)
Applying the complex-algebraic identity to the original g2 yields
![{\displaystyle g_{2}=-{3 \over 2}\left(\mathrm {Re} (z'')\mathrm {Re} (1+z^{4})-\mathrm {Im} (z'')\mathrm {Im} (1+z^{4})\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a0a8832641b775ba4f34e64a6bc926515390af2)
![{\displaystyle g_{2}=-{3 \over 2}\left[\mathrm {Re} (z'')(1+\mathrm {Re} (z^{4}))-\mathrm {Im} (z'')\mathrm {Im} (z^{4})\right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bdda93c035e68eba4f3ea2503902bd5ee6c7e834)
![{\displaystyle g_{2}=-{3 \over 2}\left[\mathrm {Re} (z'')+\mathrm {Re} (z'')\mathrm {Re} (z^{4})-\mathrm {Im} (z'')\mathrm {Im} (z^{4})\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/033f29ec3e1b149ccd89cd079c358d3f04a526d4)
The raw coordinates of the pre-rotated point are
![{\displaystyle g_{1}=-{3 \over 2}[\mathrm {Im} (z'')-\mathrm {Im} (z'')\mathrm {Re} (z^{4})-\mathrm {Re} (z'')\mathrm {Im} (z^{4})],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44cd0e094efe85729029ac422b4407a9d020d198)
![{\displaystyle g_{2}=-{3 \over 2}\left[\mathrm {Re} (z'')+\mathrm {Re} (z'')\mathrm {Re} (z^{4})-\mathrm {Im} (z'')\mathrm {Im} (z^{4})\right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8135e27fd4cc4004c2840f97cbcebf150e9f687)
and the raw coordinates of the post-rotated point are
![{\displaystyle g_{1}'=-{3 \over 2}\left[-{1 \over 2}\mathrm {Im} (z'')+{1 \over 2}\mathrm {Im} (z'')\mathrm {Re} (z^{4})-{{\sqrt {3}} \over 2}\mathrm {Im} (z'')\mathrm {Im} (z^{4})+{{\sqrt {3}} \over 2}\mathrm {Re} (z'')+{1 \over 2}\mathrm {Re} (z'')\mathrm {Im} (z^{4})+{{\sqrt {3}} \over 2}\mathrm {Re} (z'')\mathrm {Re} (z^{4})\right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/673939576e991f607d5df49b1a9a5f39d940404b)
![{\displaystyle g_{2}'=-{3 \over 2}\left[-{1 \over 2}\mathrm {Re} (z'')-{1 \over 2}\mathrm {Re} (z'')\mathrm {Re} (z^{4})+{{\sqrt {3}} \over 2}\mathrm {Re} (z'')\mathrm {Im} (z^{4})-{{\sqrt {3}} \over 2}\mathrm {Im} (z'')+{{\sqrt {3}} \over 2}\mathrm {Im} (z'')\mathrm {Re} (z^{4})+{1 \over 2}\mathrm {Im} (z'')\mathrm {Im} (z^{4})\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09aaa064f48fafb2a654719b612c32c7efbc703e)
Comparing these four coordinates we can verify that
![{\displaystyle g_{1}'=-{1 \over 2}g_{1}+{{\sqrt {3}} \over 2}g_{2},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1208262f38e9bc8091aed02ba860f10b8bf19363)
![{\displaystyle g_{2}'=-{{\sqrt {3}} \over 2}g_{1}-{1 \over 2}g_{2}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e35d7fff0750b70d555ac22d6f9dfeee5a9cc878)
In matrix form, this can be expressed as
![{\displaystyle {\begin{bmatrix}g_{1}'\\g_{2}'\\g_{3}'\end{bmatrix}}={\begin{bmatrix}-{1 \over 2}&{{\sqrt {3}} \over 2}&0\\-{{\sqrt {3}} \over 2}&-{1 \over 2}&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}g_{1}\\g_{2}\\g_{3}\end{bmatrix}}={\begin{bmatrix}\cos {-2\pi \over 3}&-\sin {-2\pi \over 3}&0\\\sin {-2\pi \over 3}&\cos {-2\pi \over 3}&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}g_{1}\\g_{2}\\g_{3}\end{bmatrix}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5711de9557f17402e2ce90fa8daabc1fdb6924a)
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.