If z is replaced by the negative reciprocal of its complex conjugate ,
−
1
z
⋆
,
{\displaystyle -{1 \over z^{\star }},}
g1 , g2 , and g3 of z are left unchanged.
Let g1 ′ be obtained from g1 by substituting z with
−
1
z
⋆
.
{\displaystyle -{1 \over z^{\star }}.}
g
1
′
=
−
3
2
I
m
(
−
1
z
⋆
(
1
−
1
z
⋆
4
)
1
z
⋆
6
−
5
1
z
⋆
3
−
1
)
.
{\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).}
Multiply both numerator and denominator by
z
⋆
6
,
{\displaystyle z^{\star 6},}
g
1
′
=
−
3
2
I
m
(
−
z
⋆
(
z
⋆
4
−
1
)
1
−
5
z
⋆
3
−
z
⋆
6
)
.
{\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).}
Multiply both numerator and denominator by -1,
g
1
′
=
−
3
2
I
m
(
z
⋆
(
z
⋆
4
−
1
)
z
⋆
6
+
5
z
⋆
3
−
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).}
It is generally true for any complex number z and any integral power n that
(
z
⋆
)
n
=
(
z
n
)
⋆
,
{\displaystyle (z^{\star })^{n}=(z^{n})^{\star },}
therefore
g
1
′
=
−
3
2
I
m
(
z
⋆
(
z
4
−
1
)
⋆
(
z
6
+
5
z
3
−
1
)
⋆
)
,
{\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),}
g
1
′
=
−
3
2
I
m
(
−
(
z
(
1
−
z
4
)
z
6
+
5
z
3
−
1
)
⋆
)
{\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)}
therefore
g
1
′
=
g
1
{\displaystyle g_{1}'=g_{1}}
complex number z ,
I
m
(
−
z
⋆
)
=
I
m
(
z
)
.
{\displaystyle \mathrm {Im} (-z^{\star })=\mathrm {Im} (z).}
Let g2 ′ be obtained from g2 by substituting z with
−
1
z
⋆
.
{\displaystyle -{1 \over z^{\star }}.}
g
2
′
=
−
3
2
R
e
(
−
1
z
⋆
(
1
+
1
z
⋆
4
)
1
z
⋆
6
−
5
1
z
⋆
3
−
1
)
,
{\displaystyle g_{2}'=-{3 \over 2}\mathrm {Re} \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),}
=
−
3
2
R
e
(
z
⋆
(
z
⋆
4
+
1
)
z
⋆
6
+
5
z
⋆
3
−
1
)
,
{\displaystyle =-{3 \over 2}\mathrm {Re} \left({z^{\star }(z^{\star 4}+1) \over z^{\star 6}+{\sqrt {5}}z^{\star 3}-1}\right),}
=
−
3
2
R
e
(
z
⋆
(
z
4
⋆
+
1
)
z
6
⋆
+
5
z
3
⋆
−
1
)
,
{\displaystyle =-{3 \over 2}\mathrm {Re} \left({z^{\star }(z^{4\star }+1) \over z^{6\star }+{\sqrt {5}}z^{3\star }-1}\right),}
=
−
3
2
R
e
(
z
⋆
(
z
4
+
1
)
⋆
(
z
6
+
5
z
3
−
1
)
⋆
)
,
{\displaystyle =-{3 \over 2}\mathrm {Re} \left({z^{\star }(z^{4}+1)^{\star } \over (z^{6}+{\sqrt {5}}z^{3}-1)^{\star }}\right),}
=
−
3
2
R
e
(
(
z
(
z
4
+
1
)
z
6
+
5
z
3
−
1
)
⋆
)
,
{\displaystyle =-{3 \over 2}\mathrm {Re} \left(\left({z(z^{4}+1) \over z^{6}+{\sqrt {5}}z^{3}-1}\right)^{\star }\right),}
therefore
g
2
′
=
g
2
{\displaystyle g_{2}'=g_{2}}
z ,
R
e
(
z
⋆
)
=
R
e
(
z
)
.
{\displaystyle \mathrm {Re} (z^{\star })=\mathrm {Re} (z).}
Let g3 ′ be obtained from g3 by substituting z with
−
1
z
⋆
.
{\displaystyle -{1 \over z^{\star }}.}
g
3
′
=
I
m
(
1
+
1
z
⋆
6
1
z
⋆
6
−
5
1
z
⋆
3
−
1
)
,
{\displaystyle g_{3}'=\mathrm {Im} \left({1+{1 \over z^{\star 6}} \over {1 \over z^{\star 6}}-{\sqrt {5}}{1 \over z^{\star 3}}-1}\right),}
=
I
m
(
z
⋆
6
+
1
1
−
5
z
⋆
3
−
z
⋆
6
)
,
{\displaystyle =\mathrm {Im} \left({z^{\star 6}+1 \over 1-{\sqrt {5}}z^{\star 3}-z^{\star 6}}\right),}
=
I
m
(
z
6
⋆
+
1
1
−
5
z
3
⋆
−
z
6
⋆
)
,
{\displaystyle =\mathrm {Im} \left({z^{6\star }+1 \over 1-{\sqrt {5}}z^{3\star }-z^{6\star }}\right),}
=
I
m
(
−
(
z
6
+
1
)
⋆
(
z
6
+
5
z
3
−
1
)
⋆
)
,
{\displaystyle =\mathrm {Im} \left(-{(z^{6}+1)^{\star } \over (z^{6}+{\sqrt {5}}z^{3}-1)^{\star }}\right),}
=
I
m
(
−
(
z
6
+
1
z
6
+
5
z
3
−
1
)
⋆
)
,
{\displaystyle =\mathrm {Im} \left(-\left({z^{6}+1 \over z^{6}+{\sqrt {5}}z^{3}-1}\right)^{\star }\right),}
therefore
g
3
′
=
g
3
.
{\displaystyle g_{3}'=g_{3}.}
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
R
e
(
U
V
)
=
R
e
(
U
)
R
e
(
V
)
−
I
m
(
U
)
I
m
(
V
)
,
{\displaystyle \mathrm {Re} (UV)=\mathrm {Re} (U)\mathrm {Re} (V)-\mathrm {Im} (U)\mathrm {Im} (V),\,\!}
I
m
(
U
V
)
=
R
e
(
U
)
I
m
(
V
)
+
I
m
(
U
)
R
e
(
V
)
.
{\displaystyle \mathrm {Im} (UV)=\mathrm {Re} (U)\mathrm {Im} (V)+\mathrm {Im} (U)\mathrm {Re} (V).\,\!}
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
z
′
=
z
e
i
2
π
/
3
{\displaystyle z'=ze^{i2\pi /3}\,\!}
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
g
3
′
(
z
′
)
=
I
m
(
1
+
z
′
6
z
′
6
+
5
z
′
3
−
1
)
−
1
2
,
{\displaystyle g_{3}'(z')=\mathrm {Im} \left({1+z'^{6} \over z'^{6}+{\sqrt {5}}z'^{3}-1}\right)-{1 \over 2},}
g
3
′
(
z
)
=
I
m
(
1
+
z
6
e
i
4
π
z
6
e
i
4
π
+
5
z
i
2
π
−
1
)
−
1
2
.
{\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}.}
Since
e
i
4
π
=
e
i
2
π
=
1
,
{\displaystyle e^{i4\pi }=e^{i2\pi }=1,}
g
3
′
=
I
m
(
1
+
z
6
z
6
+
5
z
3
−
1
)
−
1
2
{\displaystyle g_{3}'=\mathrm {Im} \left({1+z^{6} \over z^{6}+{\sqrt {5}}z^{3}-1}\right)-{1 \over 2}}
therefore
g
3
′
=
g
3
.
{\displaystyle g_{3}'=g_{3}.}
Z -axis.
Plug in z′ for z in g1 (z) , resulting in
g
1
′
(
z
)
=
−
3
2
I
m
(
z
e
i
2
π
/
3
(
1
−
z
4
e
i
8
π
/
3
)
z
6
e
i
4
π
+
5
z
3
e
i
2
π
−
1
)
.
{\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).}
Noticing that
e
i
8
π
/
3
=
e
i
2
π
/
3
,
{\displaystyle e^{i8\pi /3}=e^{i2\pi /3},}
g
1
′
=
−
3
2
I
m
(
z
e
i
2
π
/
3
(
1
−
z
4
e
i
2
π
/
3
)
z
6
+
5
z
3
−
1
)
.
{\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).}
Then, letting
e
i
4
π
/
3
=
e
−
i
2
π
/
3
{\displaystyle e^{i4\pi /3}=e^{-i2\pi /3}}
g
1
′
=
−
3
2
I
m
(
z
(
e
i
2
π
/
3
−
z
4
e
−
i
2
π
/
3
)
z
6
+
5
z
3
−
1
)
.
{\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).}
Now, applying the complex-algebraic identity, and letting
z
″
=
z
z
6
+
5
z
3
−
1
{\displaystyle z''={z \over z^{6}+{\sqrt {5}}z^{3}-1}}
we get
g
1
′
=
−
3
2
[
I
m
(
z
″
)
R
e
(
e
i
2
π
/
3
−
z
4
e
−
i
2
π
/
3
)
+
R
e
(
z
″
)
I
m
(
e
i
2
π
/
3
−
z
4
e
−
i
2
π
/
3
)
]
.
{\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].}
Both
R
e
{\displaystyle \mathrm {Re} }
I
m
{\displaystyle \mathrm {Im} }
distributive with respect to addition, and
R
e
(
e
i
θ
)
=
cos
θ
,
{\displaystyle \mathrm {Re} (e^{i\theta })=\cos \theta ,\,\!}
I
m
(
e
i
θ
)
=
sin
θ
,
{\displaystyle \mathrm {Im} (e^{i\theta })=\sin \theta ,\,\!}
due to Euler's formula , so that
g
1
′
=
−
3
2
[
I
m
(
z
″
)
(
cos
2
π
3
−
R
e
(
z
4
e
−
i
2
π
/
3
)
)
+
R
e
(
z
″
)
(
sin
2
π
3
−
I
m
(
z
4
e
−
i
2
π
/
3
)
)
]
.
{\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].}
Applying the complex-algebraic identities again, and simplifying
cos
2
π
3
{\displaystyle \cos {2\pi \over 3}}
sin
2
π
3
{\displaystyle \sin {2\pi \over 3}}
3
/
2
,
{\displaystyle {\sqrt {3}}/2,}
g
1
′
=
−
3
2
[
I
m
(
z
″
)
(
−
1
2
−
[
R
e
(
z
4
)
R
e
(
e
−
i
2
π
/
3
)
−
I
m
(
z
4
)
I
m
(
e
−
i
2
π
/
3
)
]
)
+
R
e
(
z
″
)
(
3
2
−
[
I
m
(
z
4
)
R
e
(
e
−
i
2
π
/
3
)
+
R
e
(
z
4
)
I
m
(
e
−
i
2
π
/
3
)
]
)
]
.
{\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].}
Simplify constants,
g
1
′
=
−
3
2
[
I
m
(
z
″
)
(
−
1
2
−
[
−
1
2
R
e
(
z
4
)
+
3
2
I
m
(
z
4
)
]
)
+
R
e
(
z
″
)
(
3
2
−
[
−
1
2
I
m
(
z
4
)
−
3
2
R
e
(
z
4
)
]
)
]
,
{\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],}
therefore
g
1
′
=
−
3
2
[
−
1
2
I
m
(
z
″
)
+
1
2
I
m
(
z
″
)
R
e
(
z
4
)
−
3
2
I
m
(
z
″
)
I
m
(
z
4
)
+
3
2
R
e
(
z
″
)
+
1
2
R
e
(
z
″
)
I
m
(
z
4
)
+
3
2
R
e
(
z
″
)
R
e
(
z
4
)
]
.
{\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].}
Applying the complex-algebraic identity to the original g1 yields
g
1
=
−
3
2
[
I
m
(
z
″
)
R
e
(
1
−
z
4
)
+
R
e
(
z
″
)
I
m
(
1
−
z
4
)
]
,
{\displaystyle g_{1}=-{3 \over 2}[\mathrm {Im} (z'')\mathrm {Re} (1-z^{4})+\mathrm {Re} (z'')\mathrm {Im} (1-z^{4})],}
g
1
=
−
3
2
[
I
m
(
z
″
)
(
1
−
R
e
(
z
4
)
)
+
R
e
(
z
″
)
(
−
I
m
(
z
4
)
)
]
,
{\displaystyle g_{1}=-{3 \over 2}[\mathrm {Im} (z'')(1-\mathrm {Re} (z^{4}))+\mathrm {Re} (z'')(-\mathrm {Im} (z^{4}))],}
g
1
=
−
3
2
[
I
m
(
z
″
)
−
I
m
(
z
″
)
R
e
(
z
4
)
−
R
e
(
z
″
)
I
m
(
z
4
)
]
.
{\displaystyle g_{1}=-{3 \over 2}[\mathrm {Im} (z'')-\mathrm {Im} (z'')\mathrm {Re} (z^{4})-\mathrm {Re} (z'')\mathrm {Im} (z^{4})].}
Plug in z′ for z in g2 (z) , resulting in
g
2
′
=
−
3
2
R
e
(
z
e
i
2
π
/
3
(
1
+
z
4
e
i
8
π
/
3
)
z
6
e
i
4
π
+
5
z
3
e
i
2
π
−
1
)
.
{\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).}
Simplify the exponents,
g
2
′
=
−
3
2
R
e
(
z
e
i
2
π
/
3
(
1
+
z
4
e
i
2
π
/
3
)
z
6
+
5
z
3
−
1
)
,
{\displaystyle g_{2}'=-{3 \over 2}\mathrm {Re} \left({ze^{i2\pi /3}(1+z^{4}e^{i2\pi /3}) \over z^{6}+{\sqrt {5}}z^{3}-1}\right),}
=
−
3
2
R
e
(
z
″
(
e
i
2
π
/
3
+
z
4
e
−
i
2
π
/
3
)
)
.
{\displaystyle =-{3 \over 2}\mathrm {Re} (z''(e^{i2\pi /3}+z^{4}e^{-i2\pi /3})).}
Now apply the complex-algebraic identity to g′2 , obtaining
g
2
′
=
−
3
2
[
R
e
(
z
″
)
R
e
(
e
i
2
π
/
3
+
z
4
e
−
i
2
π
/
3
)
−
I
m
(
z
″
)
I
m
(
e
i
2
π
/
3
+
z
4
e
−
i
2
π
/
3
)
]
.
{\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].}
Distribute the
R
e
{\displaystyle \mathrm {Re} }
g
2
′
=
−
3
2
[
R
e
(
z
″
)
(
−
1
2
+
R
e
(
z
4
e
−
i
2
π
/
3
)
)
−
I
m
(
z
″
)
(
3
2
+
I
m
(
z
4
e
−
i
2
π
/
3
)
)
]
.
{\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].}
Apply the complex-algebraic identities again,
g
2
′
=
−
3
2
[
R
e
(
z
″
)
(
−
1
2
+
R
e
(
z
4
)
R
e
(
e
−
i
2
π
/
3
)
−
I
m
(
z
4
)
I
m
(
e
−
i
2
π
3
)
)
−
I
m
(
z
″
)
(
3
2
+
I
m
(
z
4
)
R
e
(
e
−
i
2
π
/
3
)
+
R
e
(
z
4
)
I
m
(
e
−
i
2
π
/
3
)
)
]
.
{\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].}
Simplify constants,
g
2
′
=
−
3
2
[
R
e
(
z
″
)
(
−
1
2
−
1
2
R
e
(
z
4
)
+
3
2
I
m
(
z
4
)
)
−
I
m
(
z
″
)
(
3
2
−
3
2
R
e
(
z
4
)
−
1
2
I
m
(
z
4
)
)
]
,
{\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],}
then distribute with respect to addition,
g
2
′
=
−
3
2
[
−
1
2
R
e
(
z
″
)
−
1
2
R
e
(
z
″
)
R
e
(
z
4
)
+
3
2
R
e
(
z
″
)
I
m
(
z
4
)
−
3
2
I
m
(
z
″
)
+
3
2
I
m
(
z
″
)
R
e
(
z
4
)
+
1
2
I
m
(
z
″
)
I
m
(
z
4
)
]
.
{\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].}
Applying the complex-algebraic identity to the original g2 yields
g
2
=
−
3
2
(
R
e
(
z
″
)
R
e
(
1
+
z
4
)
−
I
m
(
z
″
)
I
m
(
1
+
z
4
)
)
,
{\displaystyle g_{2}=-{3 \over 2}\left(\mathrm {Re} (z'')\mathrm {Re} (1+z^{4})-\mathrm {Im} (z'')\mathrm {Im} (1+z^{4})\right),}
g
2
=
−
3
2
[
R
e
(
z
″
)
(
1
+
R
e
(
z
4
)
)
−
I
m
(
z
″
)
I
m
(
z
4
)
]
,
{\displaystyle g_{2}=-{3 \over 2}\left[\mathrm {Re} (z'')(1+\mathrm {Re} (z^{4}))-\mathrm {Im} (z'')\mathrm {Im} (z^{4})\right],}
g
2
=
−
3
2
[
R
e
(
z
″
)
+
R
e
(
z
″
)
R
e
(
z
4
)
−
I
m
(
z
″
)
I
m
(
z
4
)
]
.
{\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].}
The raw coordinates of the pre-rotated point are
g
1
=
−
3
2
[
I
m
(
z
″
)
−
I
m
(
z
″
)
R
e
(
z
4
)
−
R
e
(
z
″
)
I
m
(
z
4
)
]
,
{\displaystyle g_{1}=-{3 \over 2}[\mathrm {Im} (z'')-\mathrm {Im} (z'')\mathrm {Re} (z^{4})-\mathrm {Re} (z'')\mathrm {Im} (z^{4})],}
g
2
=
−
3
2
[
R
e
(
z
″
)
+
R
e
(
z
″
)
R
e
(
z
4
)
−
I
m
(
z
″
)
I
m
(
z
4
)
]
,
{\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],}
and the raw coordinates of the post-rotated point are
g
1
′
=
−
3
2
[
−
1
2
I
m
(
z
″
)
+
1
2
I
m
(
z
″
)
R
e
(
z
4
)
−
3
2
I
m
(
z
″
)
I
m
(
z
4
)
+
3
2
R
e
(
z
″
)
+
1
2
R
e
(
z
″
)
I
m
(
z
4
)
+
3
2
R
e
(
z
″
)
R
e
(
z
4
)
]
,
{\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],}
g
2
′
=
−
3
2
[
−
1
2
R
e
(
z
″
)
−
1
2
R
e
(
z
″
)
R
e
(
z
4
)
+
3
2
R
e
(
z
″
)
I
m
(
z
4
)
−
3
2
I
m
(
z
″
)
+
3
2
I
m
(
z
″
)
R
e
(
z
4
)
+
1
2
I
m
(
z
″
)
I
m
(
z
4
)
]
.
{\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].}
Comparing these four coordinates we can verify that
g
1
′
=
−
1
2
g
1
+
3
2
g
2
,
{\displaystyle g_{1}'=-{1 \over 2}g_{1}+{{\sqrt {3}} \over 2}g_{2},}
g
2
′
=
−
3
2
g
1
−
1
2
g
2
.
{\displaystyle g_{2}'=-{{\sqrt {3}} \over 2}g_{1}-{1 \over 2}g_{2}.}
In matrix form, this can be expressed as
[
g
1
′
g
2
′
g
3
′
]
=
[
−
1
2
3
2
0
−
3
2
−
1
2
0
0
0
1
]
[
g
1
g
2
g
3
]
=
[
cos
−
2
π
3
−
sin
−
2
π
3
0
sin
−
2
π
3
cos
−
2
π
3
0
0
0
1
]
[
g
1
g
2
g
3
]
.
{\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}}.}
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
![{\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)
![{\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)
![{\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)
![{\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)
![{\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)
![{\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)
![{\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)
![{\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)
![{\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)
![{\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)
![{\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)
![{\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)
![{\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)
![{\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)
