An archived instance of discourse for discussion in undergraduate Complex Dynamics.

# Find your personal cubic Julia set

mark

(20 pts)

Locate your personal cubic Julia set in the parameter space of cubic Julia sets. That is:

• Find parameter values $a$ and $b$ so that
$$f_{a,b}(z) = z^3 - 3\,a^2 z + b$$
is affinely conjugate to your personal cubic polynomial.

• Open this notebook on the cubic parameter space from our class webpage and use the parameterPic command defined there to plot a slice of the cubic parameter space containing your parameters and mark the precise parameter with a point.

RedCrayon

Are we posting a slice with $a$ constant or with $b$ constant?

mark

@RedCrayon I think that either an $a$-slice or a $b$-slice would be sufficient.

RedCrayon

Let $$g(z)=(2760 + 4/10) - (8164 + 8/10) z + (8051 + 4/10) z^2 - 2646 z^3$$ be my personal cubic. For $f_{a,b}(z)$ to be affinely conjugate to $g(z)$ there must exist $c,d\in\mathbb{C}$ such that $$(f_{a,b}\circ\phi)(z)=(\phi\circ g)(z)$$ where $\phi(z)=cz+d$. Define the functions in Mathematica using

g[z_] := (2760 + 4/10) - (8164 + 8/10) z + (8051 + 4/10) z^2 - 2646 z^3
f[a_, b_][z_] := z^3 - 3 a^2 z + b
\[Phi][z_] := c z + d

Next, define a list of equations

eqs = Thread[CoefficientList[\[Phi][g[z]] - f[a, b][\[Phi][z]], z] == {0, 0, 0, 0}]

and solve using

Solve[eqs, {a, b, c, d}]

I chose the solution where $a=\frac{3i}{5}\sqrt{\frac{3}{2}}$ and $b=\frac{3i}{125}\sqrt{6}$. If we fix $b$ and take a slice of the parameter space using the parameterPic command, we get the image

where the point $a$ is marked in yellow. The code for the image above is

a = 3/5 I Sqrt[3/2]
b = (3 I Sqrt[6])/125
Show[
parameterPic[
"b", b, -1.4 - 1.4 I, 1.4 + 1.4 I, 0.005,
Frame -> True, PlotRangePadding -> None
],
ListPlot[{{Re[a], Im[a]}}, PlotStyle -> Yellow]
]

The Julia set for $f_{a,b}(z)$ with my choices for $a$ and $b$ is

The Julia set for $g(z)$ is

audrey

My personal cubic polynomial is:
$$g(z) = -531 z^3 + (48321/10) z^2 - (146556/10) z + 148179/10.$$
I wrote it as $g(z)$, rather than $f(z)$ so as not to confuse it with
$$f_{a,b}(z) = z^3 - 3a^2 z + b.$$
I guess we need $\varphi(z) = m z + c$ such that $\varphi\circ g = f_{a,b}\circ \varphi$. In Mathematica code, we need to solve the following equations:

g[z_] = -531 z^3 + (48321/10) z^2 - (146556/10) z + 148179/10;
f[a_, b_][z_] = z^3 - 3 a^2*z + b;
phi[z_] = m*z + c;
eqs = CoefficientList[phi[g[z]] - f[a, b][phi[z]], z] == {0, 0, 0, 0};
eqs // InputForm

(* Out:
{-b + c + 3*a^2*c - c^3 + (148179*m)/10,
(-73278*m)/5 + 3*a^2*m - 3*c^2*m,
(48321*m)/10 - 3*c*m^2,
-531*m - m^3} == {0, 0, 0, 0}
*)

We can solve these equations as follows:

sols = Solve[eqs, {a, b, m, c}]

(* Out:
{a -> -((I Sqrt[59])/10), b -> (9 I Sqrt[59])/500, m -> 3 I Sqrt[59], c -> -((91 I Sqrt[59])/10)},
{a -> (I Sqrt[59])/10, b -> (9 I Sqrt[59])/500, m -> 3 I Sqrt[59], c -> -((91 I Sqrt[59])/10)},
{a -> -((I Sqrt[59])/10), b -> -((9 I Sqrt[59])/500), m -> -3 I Sqrt[59], c -> (91 I Sqrt[59])/10},
{a -> (I Sqrt[59])/10, b -> -((9 I Sqrt[59])/500), m -> -3 I Sqrt[59], c -> (91 I Sqrt[59])/10}
*)

Thus, I guess my personal cubic polynomial is affinely conjugate to $f_{a,b}$, where $a$ and $b$ are given by any of those four solutions. If I choose the second one, with no leading minus signs, I get

ff[z_] = f[a, b][z] /. sols[[2]]
(* Out: (9 I Sqrt[59])/500 + (177 z)/100 + z^3 *)

Let's check the two Julia sets:

Grid[{{
JuliaSetPlot[ff[z], z, ImageSize -> {Automatic, 400}],
JuliaSetPlot[g[z], z, ImageSize -> 400,
PlotRange -> {{2.95, 3.12}, {-0.03, 0.03}}]
}}, Spacings -> 5]

Now, I need to locate this in the cubic parameter space using the parameterPic command.

parameterPic["a", (I Sqrt[59])/10,  -1 - I, 1 + I, 0.005,
Frame -> True, PlotRangePadding -> None,
Epilog -> {Yellow, PointSize[Large], Point[{0, (9 Sqrt[59])/500}]}]

SomeCallMeTim

Our personal cubic polynomial is $P_{Tim}(z)=-513z^3+4668.3z^2-14158.8z+14315.7$. We wish to find parameter values $a$ and $b$ so that the polynomial $f_{a,b}(z)=z^3-3a^2z+b$ is affinely conjugate to $P_{Tim}(z)$.

Using the mentioned notebook as a reference, we modify the code to calculate solutions for our parameters.

phi[z_] = m*z + d;
f[a_, b_][z_] = z^3 - 3 a^2*z + b;
g[z_] = -513*z^3 + 4668.3*z^2 + -14158.8*z + 14315.7;
CoefficientList[phi[g[z]] - f[a, b][phi[z]], z] == {0, 0, 0, 0}]
Solve[eqs, {a, b, d, m}]

This gives us several solutions, we select $$a=0.7549834435270749i,\; b=-0.10569768209379049i$$ $$d=68.70349336096382i,\; m=-22.64950330581225i.$$

Thus, we have $$f(z)=z^3-3(0.754983i)^2z+(-.105698209379049i)=z^3+1.71z-.105698209379049i$$
which is conjugate to $P_{Tim}(z)$ by the conjugacy $$\phi(z)=(-22.64950330581225i) z +68.70349336096382i.$$

We look back at our previous consideration of our personal julia set, and compare that image to the following image, generated in mathematica with our new $f(z)$.

pic = JuliaSetPlot[f[z], z, ImageResolution -> 1000,
PlotRange -> {{2.95, 3.11}, {-0.03, 0.03}},
Epilog -> {
{PointSize[Medium], Green, Point[{3.0000000000036087, 0}]},
{PointSize[Medium], Red, Point[{3.02653, 0}]},
{PointSize[Medium], White, Point[{3.07347, 0}]}
}
]

cubicJuliaPic[0.7549834435270749*I, -0.10569768209379049*I, 100, 600]

We note that their shape appears the same, aside from a shift, stretch, and rotation, as expected for conjugate functions.

We are now asked to utilize the parameterPic command to plot a slice of the cubic parameter space corresponding to our $a$ and $b$, with a plot marking the precise value of the varied parameter.

We first produce a fixed $a$ slice, and mark our particular $b$:

parameterPic["a", 0.7549834435270749*I, -2 - I, 2 + I, 0.005,
Frame -> True, PlotRangePadding -> None,
Epilog -> {{PointSize[Medium], Yellow,
Point[{{0, -0.10569768209379049}}]}}]

Next, we fix our $b$ value, and mark our particular value of $a$.

parameterPic["b", -0.10569768209379049*I, -2 - I, 2 + I, 0.005,
Frame -> True, PlotRangePadding -> None,
Epilog -> {{PointSize[Medium], Yellow,
Point[{{0, 0.7549834435270749}}]}}]

Person

My function is

$$g(z)=2842x3+17173.8x2−34591.2x+23225.2$$

We want to find $\phi(z)=mz+c$ such that $\phi\circ g=f_{a,b}\circ\phi$ where $f{a,b}=z^3-3a^2z+b$. So we will use the code

g[z_] = \[Minus]2842 z^3 + 171738/10 z^2 \[Minus] 345912/10 z + 232252/10;
f[a_, b_][z_] = z^3 - 3 a^2*z + b;
phi[z_] = m*z + c;
eqs = CoefficientList[phi[g[z]] - f[a, b][phi[z]], z] == {0, 0, 0, 0};
eqs // InputForm
sols = Solve[eqs, {a, b, m, c}]

to get a list of coefficents which is
\begin{align*}
b \rightarrow -c (-1 - 3 a^2 + c^2),\ m \rightarrow 0, a \rightarrow -(1/5)i \sqrt{29/2}\\
My function is

$$g(z)=2842x3+17173.8x2−34591.2x+23225.2$$

We want to find $\phi(z)=mz+c$ such that $\phi\circ g=f_{a,b}\circ\phi$ where $f{a,b}=z^3-3a^2z+b$. So we will use the code

g[z_] = \[Minus]2842 z^3 + 171738/10 z^2 \[Minus] 345912/10 z + 232252/10;
f[a_, b_][z_] = z^3 - 3 a^2*z + b;
phi[z_] = m*z + c;
eqs = CoefficientList[phi[g[z]] - f[a, b][phi[z]], z] == {0, 0, 0, 0};
eqs // InputForm
sols = Solve[eqs, {a, b, m, c}]

to get

{{b -> -c (-1 - 3 a^2 + c^2), m -> 0}, {a -> -(1/5) I Sqrt[29/2],
b -> (2 I Sqrt[58])/125, m -> 7 I Sqrt[58],
c -> -(141/5) I Sqrt[29/2]}, {a -> 1/5 I Sqrt[29/2],
b -> (2 I Sqrt[58])/125, m -> 7 I Sqrt[58],
c -> -(141/5) I Sqrt[29/2]}, {a -> -(1/5) I Sqrt[29/2],
b -> -((2 I Sqrt[58])/125), m -> -7 I Sqrt[58],
c -> 141/5 I Sqrt[29/2]}, {a -> 1/5 I Sqrt[29/2],
b -> -((2 I Sqrt[58])/125), m -> -7 I Sqrt[58],
c -> 141/5 I Sqrt[29/2]}

We then use the following command to pull out the most interesting coefficients and name them.

ff[z_] = f[a, b][z] /. sols[[2]]
b = 2 I Sqrt[58]/125

Now we will use

Grid[{{JuliaSetPlot[ff[z], z, ImageSize -> {Automatic, 400}], JuliaSetPlot[g[z], z]}}, Spacings -> 5]

to get

Now we will find $g(z)$ in the cubic parameter space by using the command

parameterPic["a", b, -1.5 - 1.5 I, 1.5 + 1.5 I, 0.005, Frame -> True,
PlotRangePadding -> None, Epilog -> {Yellow, PointSize[Large], Point[{0, Re[b]}]}]

to get

My personal polynomial is $g(z)=63.4−172.8z+158.4z^2−48z^3.$
We need $φ(z)=mz+c$ such that $φ∘g=f_{a,b}∘φ$.
In Mathematica, I solved the following equations:

 g[z_] =63.4−172.8z+158.4z^2−48z^3;
f[a_, b_][z_] = z^3 - 3 a^2*z + b;
phi[z_] = m*z + c;
eqs = CoefficientList[phi[g[z]] - f[a, b][phi[z]], z] == {0, 0, 0, 0};
eqs // InputForm

(* Out:
{-b + c + 3*a^2*c - c^3 + 63.4*m,
-172.8*m + 3*a^2*m - 3*c^2*m,
158.4*m - 3*c*m^2, -48*m - m^3} == {0, 0, 0, 0}
*)

To solve this I used:

  sols = Solve[eqs, {a, b, m, c}]
(* Out:
{{b -> -1. c (-1. - 3. a^2 + c^2), m -> 0.},
{a -> 0. - 0.69282 I, b -> 0. - 0.0277128 I, m -> 0. + 6.9282 I, c -> 0. - 7.62102 I},
{a -> 0. + 0.69282 I, b -> 0. - 0.0277128 I, m -> 0. + 6.9282 I, c -> 0. - 7.62102 I},
{a -> 0. - 0.69282 I, b -> 0. + 0.0277128 I, m -> 0. - 6.9282 I, c -> 0. + 7.62102 I},
{a -> 0. + 0.69282 I, b -> 0. + 0.0277128 I, m -> 0. - 6.9282 I, c -> 0. + 7.62102 I}}
*)

So my personal cubic polynomial is affinely conjugate to $f_{a,b}$, where $a$ and $b$ are given by any of those four solutions. If I use the second solution with no leading minus sign to check the conjugacy, I get:

ff[z_] = f[a, b][z] /. sols[[2]]
(* Out: (0. - 0.0277128 i) + (1.44 + 0. i) z + z^3*)

To check the Julia sets I used:

 Grid[{{JuliaSetPlot[ff[z], z, ImageSize -> {Automatic, 400}],
JuliaSetPlot[g[z], z, ImageSize -> 400,
PlotRange -> {{2.95, 3.12}, {-0.03, 0.03}}]}}, Spacings -> 5]

But I could only get the vertical Julia set to appear. Does anyone know what I did wrong here to get only one solid picture instead of two? It is probably something simple that I am overlooking.

Using the parameterPic command I will locate this in the cubic parameter space and create a cool picture using the following code in Mathematica:

  parameterPic["a", 0.6928203230275509 I, -1 - I, 1 + I, 0.005,
Frame -> True, PlotRangePadding -> None,
Epilog -> {Yellow, PointSize[Large], Point[{0, (9 Sqrt[59])/500}]}]

And I got this image:

The resolution isn't the best because I had to copy and paste the image onto a new Mathematica document to get the image to save for some reason, but it still looks groovy.

Captain_Flapjack

My guess for the grid is that you need to change your plot range for the 2nd JuliaSePlot in order to see what you want to see.

Captain_Flapjack

So, following basically the same steps as everyone else, but with perhaps some slightly different naming. My personal cubic is:

$$f(z)=-441z^3-1278.9z^2-1234.8z-397.9$$

So I created two new functions $\phi(z)$ and $g_{a,b}(z)$ such that my polynomial is affinely conjugate to $g$ for some $a,b$ via $\phi$ for some $c,d$:

$$\begin{array}{rcl} g_{a,b}(z)&=&z^3-3a^2z+b\\ &\text{and}&\\ \phi(z)&=&cz+d \end{array}$$

$$(g_{a,b}\circ\phi )(z)=(\phi\circ g)(z)$$

for $a,b,c,$ and $d$:

f[z_] := -3979/10 - 12348/10 z - 12789/10 z^2 - 441 z^3
g[a_, b_][z_] := z^3 - 3 a^2*z + b
ϕ [z_] := c*z + d
eqs := CoefficientList[ϕ [f[z]] - g[a, b][ϕ [z]], z] == {0, 0,
0, 0}
eqs // InputForm;
sols = Solve[eqs, {a, b, c, d}]

(* Out:
{{b -> -d (-1 - 3 a^2 + d^2), c -> 0},
{a -> -((7 I)/10),  b -> (7 I)/500, c -> -21 I, d -> -((203 I)/10)},
{a -> (7 I)/10, b -> (7 I)/500, c -> -21 I, d -> -((203 I)/10)},
{a -> -((7 I)/10), b -> -((7 I)/500), c -> 21 I, d -> (203 I)/10},
{a -> (7 I)/10, b -> -((7 I)/500), c -> 21 I, d -> (203 I)/10}}
*)

Like many others, I chose the solution where the values of $a,b$ were both positive. For me, this was where $a=\frac{7i}{10}$ and $b=\frac{7i}{500}$. Then I compared the Julia sets and, like any others, found my conjugated Julia set to be rotated, translated, and streched, but the same shape:

ff[z_] = g[a, b][z] /. sols[[3]]
Grid[{{JuliaSetPlot[ff[z], z, ImageSize -> {Automatic, 400}],
JuliaSetPlot[f[z], z, ImageSize -> 400,
PlotRange -> {{-1.05, -.88}, {-0.03, 0.03}}]}}, Spacings -> 5]

So now I had to test my two parameterPic diagrams, one with $a$ fixed, and one with $b$ fixed

parameterPic["a", (7 I)/10, -1 - I, 1 + I, 0.005, Frame -> True,
Epilog -> {Yellow, PointSize[Large], Point[{0, 7 /500}]}]

parameterPic["b", (7 I)/500, -1 - I, 1 + I, 0.005, Frame -> True,
Epilog -> {Yellow, PointSize[Large], Point[{0, 7 /10}]}]

Levente

My personal cubic polynomial was $f(z) = 115.5 z^2 - 2695 z^3$, we search for a conjugation $\phi(z) = c*z + d$ so that $f$ is affinely conjugate to $g_{a,b}(z)$.

To do this I used the following code in Mathematica

f[z_] := 115.5 z^2 - 2695 z^3
p[z_] := c*z + d
g[a_, b_][z_] := z^3 - 3 a^2*z + b
CoefficientList[p[f[z]] - g[a, b][p[z]], z] == {0, 0, 0, 0}]
Solve[eqs, {a, b, c, d}]

One of the solutions that I found was $a = -0.74162 i, b = 0.074162 i, c = 51.9134 i, d = -0.74162 i$

This means $g_{a,b}(z) = z^3 -1.6500006732 z + .074162i$ and $\phi(z) = 51.934i*z - .74162i$

Now we backtrack to the Julia set of our original $f(z)$, utilizing Mathematica we can generate the Julia set for $f$ and $g$ and compare them

f[z_] = 115.5 z^2 - 2695 z^3;
pic = JuliaSetPlot[f[z], z, ImageResolution ->     1000,
PlotRange -> {{-.02, .05}, {-0.02, 0.02}},
Epilog -> {{PointSize[Medium], Green,
Point[{{0, 0}, {0.012041, 0}}]},     {PointSize[Medium], Red,
Point[{0.0308161, 0}]}}]

and now for $g$ we use the same code and just change the PlotRange parameters and the function to get something that does correspond to a stretch of magnitude $51.934$ a rotation of $\pi / 2$ and a shift by $-.74162 i$

Now we use the parameter pic plot command to show the diagrams with 'b' fixed and 'a' changing, in yellow we have highlighted our value of 'a' namely $a = .74162i$

parameterPic["b", -.074162 I, -1.5 - 1.5 I, 1.5 + 1.5 I, 0.005,
Frame -> True, PlotRangePadding -> None,
Epilog -> {Yellow, PointSize[Large], Point[{0, 0.74162}]}]

@mark Thanks for a good class

notneds

My personal cubic is:
$f(x) = 1220.7 - 1180.8 x + 381.3 x^2 - 41 x^3$
To find parameter values so and b so that
$f_{a,b}[z] = z^3 - 3 a^2*z + b;$
is affinity conjugate to my personal cubic we need to solve

hi[z_] = m*z + d;
f[a_, b_][z_] = z^3 - 3 a^2*z + b;
g[z_] = -41*z^3 + 381.3*z^2 - 1180.8*z + 1220.7;
eqs = Thread[CoefficientList[phi[g[z]] - f[a, b][phi[z]], z] == {0, 0, 0, 0}]

Which was taken from the mathematica notebook.
When doing so we get 4 solutions

solutions = {{b -> -1. d (-1. - 3. a^2 + d^2),
m -> 0.}, {a -> 0. - 0.6403124237432849 I,
b -> 0. - 0.11525623627379128 I, d -> 0. - 19.84968513604183 I,
m -> 0. + 6.4031242374328485 I}, {a ->
0. + 0.6403124237432849 I, b -> 0. - 0.11525623627379128 I,
d -> 0. - 19.84968513604183 I,
m -> 0. + 6.4031242374328485 I}, {a ->
0. - 0.6403124237432849 I, b -> 0. + 0.11525623627379128 I,
d -> 0. + 19.84968513604183 I,
m -> 0. - 6.4031242374328485 I}, {a ->
0. + 0.6403124237432849 I, b -> 0. + 0.11525623627379128 I,
d -> 0. + 19.84968513604183 I, m -> 0. - 6.4031242374328485 I}}

I used the fourth to get

ff[z_] = f[a, b][z] /. solutions[[4]]
out = (0. + 0.115256 I) + (1.23 + 0. I) z + z^3

which gives the Julia set

compared with the julia set of my personal cubic it looks correct!

Now to locate this in the cubic parameter space we will use

parameterPic["a", 0.6403124237432849 I, -1 - I, 1 + I, 0.005,
Frame -> True, PlotRangePadding -> None,
Epilog -> {Yellow, PointSize[Large],
Point[{0, 0.11525623627379128}]}]

which yields

WillHeDoTheAssignmen

My function is

g[z_] = -406 - 1260 z - 1305 z^2 - 450 z^3;

We want to find $\phi(z) = mz+c$ such that $\phi \circ g = f_{a,b} \circ \phi$. In mathematica, solving

phi[z_] = m*z + d;
f[a_, b_][z_] = z^3 + 3 a^2*z + b;
g[z_] = -406 - 1260 z - 1305 z^2 - 450 z^3;
CoefficientList[phi[g[z]] - f[a, b][phi[z]], z] == {0, 0, 0, 0}]
Solve[eqs, {a, b, d, m}]

This gives us a list of 4 possible coefficients each for a, b, d, and m. Choosing the combination with all positive values of $a= \frac{1}{\sqrt{2}} = 0.707106, b = 0, d = \frac{29i}{\sqrt{2}} = 20.5061i,$ and $m = 15i\sqrt{2} = 21.2132i$.

Therefore our new $f(z)$ function is $f(z) = z^3 - 3(0.707106)^2z + 0$ = $z^3 +1.5z$. This is conjugate to $\phi(z) = 21.2132i*z + 20.5061i$.

Checking the 2 Julia sets gives us

Grid[{{JuliaSetPlot[ff[z], z, ImageSize -> {Automatic, 400}],
JuliaSetPlot[g[z], z, ImageSize -> 400,
PlotRange -> {{-1.05, -.85}, {-.1, .1}}]}}, Spacings -> 5]

We can loate this in the cubic parameter space with the parameterPic command

parameterPic["a", 1/Sqrt[2], -2 - I, 2 + I, 0.005, Frame -> True,
Epilog -> {Yellow, PointSize[Large], Point[{1/Sqrt[2], 0}]}]

Rick

Lets take another look at this bad-ass cubic:
$$g(z) = 19621.6 - 29223.6 z + 14508.9 z^2 - 2401 z^3$$.
We want to find $a$ and $b$ such that $f_{a,b}(z) = z^3 - 3\,a^2 z + b$ such that $f_{a,b}$ is affinely conjugate to $f$. Thus, $f_{a,b}(\phi(z))=\phi(g(z))$ where $\phi(z)=cz+d$.
We then implement the following mathematica code:

g[z_] := 19621.6 - 29223.6 z + 14508.9 z^2 - 2401 z^3
f[a_, b_][z_] := z^3 - 3 a^2 z + b
\[Phi][z_] := c z + d
eqs = CoefficientList[\[Phi][g[z]] - f[a, b][\[Phi][z]], z] == {0, 0,
0, 0};
eqs // InputForm

Yielding,

{-b + 19621.6*c + d + 3*a^2*d - d^3, -29223.6*c + 3*a^2*c - 3*c*d^2, 14508.9*c - 3*c^2*d,
-2401*c - c^3} == {0, 0, 0, 0}

We then solve this equation using

sols = Solve[eqs, {a, b, c, d}],

yielding

{a -> 0. - 0.7 I, b -> 0. - 0.014 I, c -> 0. + 49. I, d -> 0. - 98.7 I},
{a -> 0. + 0.7 I, b -> 0. - 0.014 I, c -> 0. + 49. I, d -> 0. - 98.7 I},
{a -> 0. - 0.7 I, b -> 0. + 0.014 I, c -> 0. - 49. I, d -> 0. + 98.7 I},
{a -> 0. + 0.7 I, b -> 0. + 0.014 I, c -> 0. - 49. I, d -> 0. + 98.7 I}}

Thus, my personal cubic polynomial is affinely conjugate to $f_{a,b}$ whenever $a$ and $b$ are one of those solutions. We will now compare the julia set of the first solution with our initial cubic using the code:

h[z_] = f[a, b][z] /. sols[[2]]
Grid[{{JuliaSetPlot[h[z], z, ImageSize -> {Automatic, 400}],
JuliaSetPlot[g[z], z, ImageSize -> 400,
PlotRange -> {{1.98, 2.05}, {-0.015, 0.015}}]}}, Spacings -> 5]

Awesome. Now we can locate this is the cubic parameter space with the code:

parameterPic["a", -0.7I, -2 - I, 2 + I, 0.005, Frame -> True,
Epilog -> {Yellow, PointSize[Large], Point[{0, -0.07}]}]