imaginarypoint[x_,y_]:=Module[{m},
m=1.3*(Sqrt[(x^2)+(y^2)]);
Show[Graphics[{RGBColor[1,0,0],Line[{{0,0},{x,y},{x,0}}]}],Graphics[{RGBColor[0,0,1],Disk[{x,y},0.04*m]}],PlotRange->{{-m,m},{-m,m}},Axes->True,AxesLabel->{"Re","Im"},AspectRatio->1];
Print["θ = ",ArcTan[y/x],"    R = ",Simplify[(Sqrt[(x^2)+(y^2)])]];
]

Example 4.2.1

imaginarypoint[1,1]

Since we can have other rotations about the origin and get the same point, we can express this point in the form:

1+i=(Sqrt[2])(e^(i(π/4+2π m ))), m = ...,-2,-1,0,1,2,...

Example 4.2.3

imaginarypoint[-3,0]

Note that my little Mathematica module gets this wrong. It evaluates arctan(0)=0. However, we see that we could also have arctan(0)=π, since the following sketch reminds us that tan(π)=0. The lesson: always think about what a computer tells you!

Plot[Tan[x],{x,-2Pi,2Pi}]

Since we can have other rotations about the origin and get the same point, we can express this point in the form:

-3=3(e^(i(π+2π m ))), m = ...,-2,-1,0,1,2,...

Example 4.2.4

imaginarypoint[0,-1]

Note that my little Mathematica module gets this wrong. But we can see that θ=3π/2. This is where one of the vertical asymptotes of the tangent function is.

Plot[Tan[x],{x,-2Pi,3Pi/2}]

Since we can have other rotations about the origin and get the same point, we can express this point in the form:

-i=(e^(i(3π/2+2π m ))), m = ...,-2,-1,0,1,2,...

Example 4.2.8

imaginarypoint[1,-1]

Since we can have other rotations about the origin and get the same point, we can express this point in the form:

1-i=(Sqrt[2])(e^(i(-π/4+2π m ))), m = ...,-2,-1,0,1,2,...

An alternate form to the above, but equivalent, would be:

1-i=(Sqrt[2])(e^(i(7π/4+2π m ))), m = ...,-2,-1,0,1,2,...

Example 4.2.11

Solve[r^3-r^2-r+1==0,r]

DSolve[y'''[t]-y''[t]-y'[t]+y[t]==0,y[t],t]

Example 4.2.13

Solve[2r^3-4r^2-2r+4==0,r]

DSolve[2y'''[t]-4y''[t]-2y'[t]+4y[t]==0,y[t],t]

Example 4.2.18

Solve[(r^6)-(r^2)==0,r]

DSolve[y''''''[t]-y''[t]==0,y[t],t]

Animation: Unit Circle

imaginarypoint2[x_,y_]:=Module[{m},
m=1.3*(Sqrt[(x^2)+(y^2)]);
Show[Graphics[{RGBColor[1,0,0],Line[{{0,0},{x,y},{x,0}}]}],Graphics[{RGBColor[0,0,1],Disk[{x,y},0.04*m]}],PlotRange->{{-m,m},{-m,m}},Axes->True,AxesLabel->{"Re","Im"},AspectRatio->1]
]

Manipulate[imaginarypoint2[Cos[a],Sin[a]],{a,0,2Pi,2Pi/20}]