Manipulate Exp[I θ]+b Exp[-I θ]

Question

I've uploaded an image showing a demonstration I'd like to create, but I definitely need help. There is a lot of talent on this list, so maybe someone can help me get started learning how to add graphics to a plot and animate them.

What I'd like to create is a Manipulate demonstration that will animate the plot of Exp[I θ] + b Exp[-I θ], where b is a real number between 0 and 1. As you can see in the attached figure, I want to draw two circles centered at the origin, one with radius 1, the second with radius b. I'd like the Manipulate to have a slider that will allow me to vary the value of b between 0 and 1.

Next, I want to draw the vectors Exp[I θ] and the vector b Exp[-I θ], as pictured in the sketch. Then I want to complete the parallelogram and draw the vector Exp[I θ] + b Exp[-I θ]. I also want a slider for θthat varies from 0 to 2 π, and when I move the slider, I want to see all of the vectors move, the parallelogram adjust, and the tip of the vector Exp[I θ] + b Exp[-I θ] trace out the dotted ellipse.

I'd really appreciate it if someone could show me how to do this.

---

This was overlooked in the initial answers:
How do we add the curve Exp[I θ] + b Exp[-I θ] to the figure?

Answer 1

Manipulate[
 Module[{
   p1 = {Cos[theta ], Sin[theta]}, 
   p2 = {b Cos[-theta], b Sin[-theta ]},
   p3 = {(1 + b) Cos[theta ], (1 - b) Sin[theta ]}},
  Graphics[{
    Circle[{0, 0}],
    {Red, Circle[{0, 0}, b]},
    Arrow[{{0, 0}, p1}],
    {Red, Arrow[{{0, 0}, p2}]},
    Line[{{0, 0}, p1, p3, p2, {0, 0}}],
    {Dashed, Line[{{0, 0}, p3}]}}, 
   PlotRange -> {{-1.5, 1.5}, {-1.5, 1.5}}, Axes -> True]
  ],
 Grid[{
   {"b ", Control[{{b, .5, ""}, 0, 1, .1}], Dynamic[b]},
   {"theta ", Control[{{theta, Pi/4, ""}, 0, 2 Pi, 0.1}], 
    Row[{Dynamic[theta*180./Pi], Degree}]
    }}],
 ContinuousAction -> True
 ]

Answer 2

Blatantly copying Nasser's code, here is a slightly different style.

I used both Text and Tooltip merely to demonstrate what is possible.

Manipulate[
  With[{
    p1 = {Cos[θ], Sin[θ]},
    p2 = {b Cos[-θ], b Sin[-θ]},
    p3 = {(1 + b) Cos[θ], (1 - b) Sin[θ]}
   },
   Graphics[{
     Circle[{0, 0}],
     {Red, Circle[{0, 0}, b], Arrow[{{0, 0}, p1}], Arrow[{{0, 0}, p2}]},
     Arrow[{{0, 0}, p3}],
     {Dashed, Line[{p1, p3, p2}], Circle[{0, 0}, Abs @ {b + 1, b - 1}]},
     {Red, Text[E^(I "θ"), 1.1 p1], Text["b"/E^(I "θ"), 1.1 p2]},
     Tooltip[Point[p3], E^(I "θ") + "b"/E^(I "θ")]
    },
    PlotRange -> {{-2, 2}, {-1.3, 1.3}},
    BaseStyle -> 20,
    ImageSize -> 500
   ]
  ],
  Grid[{
    {"b ", Control[{{b, 0.5, ""}, 0, 1, 0.1}], Dynamic[b]},
    {"θ", Control[{{θ, Pi / 4, ""}, 0, 2 Pi, 0.1}], Row[{Dynamic[θ 180` / Pi], Degree}]}
  }, BaseStyle -> 20]
]

Answer 3

I used this once:

ComplexParametricPlot[f_, stuff__] := ParametricPlot[{Re[#], Im[#]} &@f, stuff]

Then putting it in the previous Manipulate (from "part I"), and giving b a complex control, we get:

Manipulate[
 Show[Graphics[
   GraphicsComplex[
    complexparts /@ {E^(I π θ), b E^(-I π θ), 
      E^(I π θ) + b E^(-I π θ), 0},
    {Circle[4],
     {Red, Circle[4, Norm[b]]}, Arrow[{4, 1}], {Red, Arrow[{4, 2}]},
     Line[{1, 3, 2}],
     {Dashed, Line[{4, 3}]}}],
   PlotRange -> 2, Axes -> True],
  ComplexParametricPlot[
   E^(I π t) + b E^(-I π t), {t, 0, θ + 10^-6}],
  ImageSize -> 250
  ],
 {{b, 0.5}, 
  LocatorPane[
    Dynamic[{Re[b], Im[b]}, (b = If[Norm[#] > 1, Complex @@ Normalize[#], Complex @@ #]) &],
    Labeled[
     Graphics[{GrayLevel[0.97], Rectangle[{-1, -1}, {1, 1}], White, 
       Disk[]}, Axes -> True, ImageSize -> 130], Row[{"b = ", Dynamic[b]}]]] &,
  ControlPlacement -> Left},
 {{θ, 1/4.}, 0., 2,
  Row[{Manipulator[##], "  ", 
     InputField[#1, Number, FieldSize -> {{0, 6}, 1}, 
      Appearance -> "Frameless"], " π"}] &},
 {{complexparts, {Re[#], Im[#]} &}, None}]

Addendum

Here's another version, inspired by Mr.Wizard, kguler, and the OP. Looks nearly the same, but it uses Circle and Rotate to trace the ellipse.

Manipulate[
 Graphics[GraphicsComplex[
   complexparts /@ {E^(I π θ), b E^(-I π θ), 
     E^(I π θ) + b E^(-I π θ), 0},
   {Circle[4],
    {Red, Circle[4, Norm[b]]}, Arrow[{4, 1}], {Red, Arrow[{4, 2}]},
    Line[{1, 3, 2}],
    {Dashed, Blue, Line[{4, 3}], 
     Rotate[
      Circle[4, Abs@{E^(I #) + b E^(-I #), E^(I #) - b E^(-I #)}, {-#, π θ - #}],
      #, {0, 0}] &[Arg[b]/2]}}],
  PlotRange -> 2, Axes -> True, ImageSize -> 250],
 {{b, 0.5 + 0.2 I}, 
  LocatorPane[
    Dynamic[{Re[b], Im[b]}, (b = If[Norm[#] > 1, Complex @@ Normalize[#], Complex @@ #]) &],
    Labeled[
     Graphics[{GrayLevel[0.97], Rectangle[{-1, -1}, {1, 1}], White, Disk[]},
      Axes -> True, ImageSize -> 130], 
     Row[{"b = ", Dynamic[b]}]]] &,
  ControlPlacement -> Left},
 {{θ, 0.25}, 0., 2.,
  Row[{Manipulator[##], "  ", 
     InputField[#1, Number, FieldSize -> {{0, 6}, 1}, 
      Appearance -> "Frameless"], " π"}] &},
 {{complexparts, {Re[#], Im[#]} &}, None}]

Answer 4

My own little modification. Deals with complex numbers E^(I θ) as such, and maps the Complex graphics to Real coordinates in a, ahem, GraphicsComplex-- sorry, can't resist a bad pun. Couldn't figure out how to get the InputField that serves as a label to format properly. (Doesn't round numbers the way a regular label will. Feel free to fix it, if you can.)

Manipulate[
 Graphics[GraphicsComplex[
   complexparts /@ {E^(I π θ), b E^(-I π θ), 
     E^(I π θ) + b E^(-I π θ), 0},
   {Circle[4],
    {Red, Circle[4, b]}, Arrow[{4, 1}], {Red, Arrow[{4, 2}]},
    Line[{1, 3, 2}],
    {Dashed, Line[{4, 3}]}}],
  PlotRange -> {{-1.5, 1.5}, {-1.5, 1.5}}, Axes -> True],
 {{b, .5}, 0, 1, Appearance -> "Labeled"},
 {{θ, 1/4.}, 0., 2,
  Row[{Manipulator[##], "  ", 
     InputField[#1, Number, FieldSize -> {{0, 6}, 1}, 
      Appearance -> "Frameless"], " π"}] &},
 {{complexparts, {Re[#], Im[#]} &}, None}]

Answer 5

Yep, the code ParametricPlot[{(1 + b) Cos[theta], (1 - b) Sin[theta]}, {theta, 0, 2 Pi}] was definitely my thought as well, but what was tough was where to put it. I finally got it to work as follows:

Manipulate[
 Module[{
   p1 = {Cos[theta], Sin[theta]},
   p2 = {b Cos[-theta], b Sin[-theta]},
   p3 = {(1 + b) Cos[theta], (1 - b) Sin[theta]}},
  Show[{
    ParametricPlot[{(1 + b) Cos[theta], (1 - b) Sin[theta]}, {theta, 
      0, 2 Pi}, PlotRange -> 1 + b],
    Graphics[{
      Circle[{0, 0}],
      {Red, Circle[{0, 0}, b]},
      Arrow[{{0, 0}, p1}],
      {Red, Arrow[{{0, 0}, p2}]},
      {Black, Dashed, Line[{{p1, p3}, {p2, p3}}]},
      {Blue, Arrow[{{0, 0}, p3}]}
      }]
    }]
  ],
 {{b, 0.5, "b"}, 0, 2},
 {{theta, 0, "theta"}, 0, 2 Pi}
 ]

I now have a working demonstration of the graph of $e^{i\theta}+be^{-i\theta}$ as $0\le\theta\le 2\pi$. There are three cases to consider: (1) $0<b<1$, (2) $b=1$, and (3) $b>1$, where $b$ is a real number. In either case, we can perform the following analysis:

$$ \begin{align*} e^{i\theta}+be^{-i\theta} &=(\cos\theta+i\sin\theta)+b(\cos(-\theta)+i\sin(-\theta))\\ &=(1+b)\cos\theta+i(1-b)\sin\theta \end{align*} $$

When we compare $z=x+iy$ with the last result, we get: $$ \begin{align*} x&=(1+b)\cos\theta\\ y&=(1-b)\sin\theta \end{align*} $$ Thus, $$ \begin{align*} \frac{x^2}{(1+b)^2}&=\cos^2\theta\\ \frac{y^2}{(1-b)^2}&=\sin^2\theta,\\ \end{align*} $$ so we can see that $$ \frac{x^2}{(1+b)^2}+\frac{x^2}{(1-b)^2}=1. $$ Thus, we have an ellipse with $x$-intercepts at $(-1-b,0)$ and $(1+b,0)$ and $y$-intercepts at $(0,-1+b)$ and $(0,1-b)$. Thus, PlotRange->1+b is a good move, as seen in the image below.

By visualizing the geometry, one can determine that the ellipse will be traced out in a counter-clockwise direction when $b<1$ and it will be traced out in a clockwise direction when $b>1$, and finally, it will trace out the line segment on the real axis between $-2$ and $2$ when $b=1$. But these cases can also be determined via the equations

$$ \begin{align*} x&=(1+b)\cos\theta\\ y&=(1-b)\sin\theta \end{align*} $$

For example, when $b=1$, these become:

$$ \begin{align*} x&=2\cos\theta\\ y&=0 \end{align*} $$

This defines the segment on the real axis between $-2$ and $2$.

There is one more step to be accomplished before I am done examining this question. I now need to select a $b$-value that is complex, that is, of the form $b=re^{i\phi}$. In this case, one can show that

$$ \begin{align*} e^{i\theta}+be^{-i\theta} &=e^{i\theta}+re^{i\phi}e^{-i\theta}\\ &=e^{i\phi}(e^{i(\theta-\phi)}+re^{-i(\theta-\phi)}), \end{align*} $$

which means that the entire image will simply be rotated through $\phi$ radians. So my final need will be to replace the $b$ slider with $r$ and $\phi$ sliders, where $r$ is a real number between 0 and 2, while $\phi$ is an angle between $-\pi/2$ and $\pi/2$. The $r$-value should affect the image in the same way as the $b$-value of the current image, while the $\phi$-value should rotate the image through $\phi$ radians.