P09-ComplexRoots.mws

High School Modules > Precalculus by Gregory A. Moore

     Roots of Complex Numbers


This worksheet goes through the development of the concept of roots of complex numbers geometrically and algebraically.

[Directions : Execute the Code Resource section first. Although there will be no output immediately, these definitions are used later in this worksheet.]

   0.  Code

>    restart; with(plots):

Warning, the name changecoords has been redefined

>    ComplexPlotY  :=  proc()
   local k,Radial,Pt,c,c1,c2,shade,r,u,Lines;
   u:= 0;
   c1 := COLOR(RGB, .67, .54, .27  );
   c2 := COLOR(RGB, .72,.6,.3 );
   for k from 1 to nargs do
       shade := evalf(rand()/10^12,2)/4 + .3;
       c := COLOR(RGB, shade+.3,shade+.2, shade);  
       Radial||k := complexplot( [0,args[k]],
              scaling = constrained, color = c1, thickness = 1);
       Lines||k := plot( [[ 0, Im(args[k])], [Re(args[k]), Im(args[k])],
                          [Re(args[k]), 0]],
                          color = c2, linestyle = 2);

       u := max( u, abs(Re(args[k])),abs(Im(args[k])) );
   od;

   r := evalf(u/12,2);
   c := COLOR(RGB, .8, .66, .25 );
   for k from 1 to nargs do
       Pt||k := plottools[disk]( [Re(args[k]), Im(args[k])],r,
                           color = c ); od;
  display( [seq( Radial||k, k = 1..nargs),  seq( Pt||k, k = 1..nargs),
            seq( Lines||k, k = 1..nargs)] );
end proc:

>    ComplexPlotB  :=  proc()
   local k,A,Pt,c,cg,shade,r,u,Lines;
   u:= 0;
   cg := COLOR(RGB, .7,.7,.8);
   for k from 1 to nargs do
       shade := evalf(rand()/10^12,2)/4 + .3;
       c := COLOR(RGB, shade,shade, shade+2);  
       A||k := complexplot( [0,args[k]], linestyle = 2,
               scaling = constrained, color = c);
       Lines||k := plot( [[ 0, Im(args[k])], [Re(args[k]), Im(args[k])],
                          [Re(args[k]), 0]], color = cg);

       u := max( u, abs(Re(args[k])),abs(Im(args[k])) );
   od;

   r := evalf(u/45,2);
   for k from 1 to nargs do
       shade := evalf(rand()/10^12,2)/5 + .1;
       c := COLOR(RGB, shade, shade, .7);
       Pt||k := plottools[rectangle](
         [Re(args[k])-r, Im(args[k])+r],[Re(args[k])+r, Im(args[k])-r],
                           color = c ); od;


  display( [seq( A||k, k = 1..nargs),  seq( Pt||k, k = 1..nargs),
            seq( Lines||k, k = 1..nargs)] );
end proc:

>    UnityRootPlot  :=  proc(n)
    local c;
     c := COLOR(RGB, .85, .65, .3  );
     display( [ComplexPlotB( 1),
          ComplexPlotY( seq( cos(2*k*Pi/n) + I*sin(2*k*Pi/n), k = 1..n) ),
          seq( plottools[arc]( [0,0], .95-k*.045, 0..(2*k*Pi/n),
                color = c,thickness = 3 ), k = 1..(n-1)),
          plottools[circle]([0,0],1, color = gold)]);
end proc:


>    UnityRoots  :=  proc(n)
    local k,S;
     S := solve(z^n - 1= 0):
     for k from 1 to n do print(S[k]); od;
end proc:

>    reset := proc()
global a,b,w,z,R,theta, phi,n,k;
a := 'a': b := 'b':  z := 'z': theta := 'theta': R := 'R':
w:='w': z:='z': phi := 'phi': n:= 'n':  k:= 'k':  
return (a,b,z,w,R,n,k,theta, phi):
end proc:

   1.  The Idea of Complex Roots


As we know there are two square roots of every real number ...

>    solve( x^2 = 9, x);

3, -3

>    3^2; (-3)^2;

9

9


Or complex numbers  ...

>    solve( x^2 = -9 );

3*I, -3*I


So it should not be surprising that there are three numbers which have same cube. In other words, 8 has three cube roots.

>    z1 := 2;    
z2 := -1 + sqrt(3)*I;  
z3 := -1 - sqrt(3)*I;

z1 := 2

z2 := -1+3^(1/2)*I

z3 := -1-I*3^(1/2)

>    z1^3: % = evalc(%);
z2^3: % = evalc(%);
z3^3: % = evalc(%);

8 = 8

(-1+3^(1/2)*I)^3 = 8

(-1-I*3^(1/2))^3 = 8

>    display(  ComplexPlotB(z1^3),
          ComplexPlotY( z1,z2,z3),
          polarplot( abs(z1), color = tan ));

[Maple Plot]

Nor should it be surprising that a number has four fourth roots.

>    z1 :=  sqrt(2) + I*sqrt(2);    
z2 :=  sqrt(2) - I*sqrt(2);
z3 := -sqrt(2) + I*sqrt(2);
z4 := -sqrt(2) - I*sqrt(2);

z1 := 2^(1/2)+2^(1/2)*I

z2 := 2^(1/2)-I*2^(1/2)

z3 := -2^(1/2)+2^(1/2)*I

z4 := -2^(1/2)-I*2^(1/2)

>    z1^4: % = evalc(%);
z2^4: % = evalc(%);
z3^4: % = evalc(%);
z4^4: % = evalc(%);

(2^(1/2)+2^(1/2)*I)^4 = -16

(2^(1/2)-I*2^(1/2))^4 = -16

(-2^(1/2)+2^(1/2)*I)^4 = -16

(-2^(1/2)-I*2^(1/2))^4 = -16

>    display(  ComplexPlotB(z1^3),
          ComplexPlotY( z1,z2,z3,z4),
          polarplot( abs(z1), color = tan ));

[Maple Plot]


Even complex numbers have four 4th roots.

>    z1 :=   .8 + 1.1*I;    
z2 :=   .8 - 1.1*I;   
z3 := - .8 + 1.1*I;   
z4 := - .8 - 1.1*I;   

z1 := .8+1.1*I

z2 := .8-1.1*I

z3 := -.8+1.1*I

z4 := -.8-1.1*I

>    z1^4: % = evalc(%);
z2^4: % = evalc(%);
z3^4: % = evalc(%);
z4^4: % = evalc(%);

-2.7727-2.0064*I = -2.7727-2.0064*I

-2.7727+2.0064*I = -2.7727+2.0064*I

-2.7727+2.0064*I = -2.7727+2.0064*I

-2.7727-2.0064*I = -2.7727-2.0064*I

>    display(  ComplexPlotB(z1^3),
          ComplexPlotY( z1,z2,z3,z4),
          polarplot( abs(z1), color = tan ));

[Maple Plot]


There is clearly some pattern to these roots, which we will explore in some depth.

   2.  Expanding One Root of Unity into Many


Notice that the number 1 raised to the 7th power is 1. Thus, it is a 7th root of 1. We would like to find more numbers which are also 7th roots of 1. In fact, there are six more.

>    phi := 2*Pi/7;

phi := 2/7*Pi

>    u := cos(phi) + I*sin(phi);
u^7:
% = round(evalf(%));

u := cos(2/7*Pi)+sin(2/7*Pi)*I

(cos(2/7*Pi)+sin(2/7*Pi)*I)^7 = 1


Given an angle like
2*Pi/7 , there is only one integer which can be multiplied to give 2*Pi .

>    phi*x = 2*Pi;
solve(%,x);

2/7*Pi*x = 2*Pi

7


However, there are other angles,  like
2*Pi/7 , that can be multiplied by 7 to give other even multiples of 2*Pi  - which are angularly equivalent to 2*Pi .

>    for k from 1 to 7 do `(7)`*k*'phi' = 7*k*phi; od;

`(7)`*phi = 2*Pi

2*`(7)`*phi = 4*Pi

3*`(7)`*phi = 6*Pi

4*`(7)`*phi = 8*Pi

5*`(7)`*phi = 10*Pi

6*`(7)`*phi = 12*Pi

7*`(7)`*phi = 14*Pi


So there are 7 different angles that can be multiplied by 7 to give an angle equivalent to 0.


Now, if we recall one of the basic properties of DeMoivre's Theorem and apply it to a number u, which is on the unit circle, we have :

>    u:='u':
u   = ( cos(theta) + I*sin(theta) );
u^n = ( cos(n*theta) + I*sin(n*theta) );

u = cos(theta)+sin(theta)*I

u^n = cos(n*theta)+sin(n*theta)*I


Now if we put these ideas together, we find 7 different complex numbers which give the same result when raised to the 7th power. Here are seven numbers.

>    phi := 2*Pi/7:
for k from 1 to 7 do
     phi||k := phi*k; od;

phi1 := 2/7*Pi

phi2 := 4/7*Pi

phi3 := 6/7*Pi

phi4 := 8/7*Pi

phi5 := 10/7*Pi

phi6 := 12/7*Pi

phi7 := 2*Pi


But this list of numbers gives the same results, since 0 and
2*Pi  are equivalent angles.

>    for k from 0 to 6 do
     phi||k := phi*k; od;

phi0 := 0

phi1 := 2/7*Pi

phi2 := 4/7*Pi

phi3 := 6/7*Pi

phi4 := 8/7*Pi

phi5 := 10/7*Pi

phi6 := 12/7*Pi


If we convert these numbers to complex numbers having these angles as their arguments, we get :

>    for k from 0 to 6 do
     cos(phi||k) + I*sin(phi||k); od;

1

cos(2/7*Pi)+sin(2/7*Pi)*I

-cos(3/7*Pi)+sin(3/7*Pi)*I

-cos(1/7*Pi)+sin(1/7*Pi)*I

-cos(1/7*Pi)-I*sin(1/7*Pi)

-cos(3/7*Pi)-I*sin(3/7*Pi)

cos(2/7*Pi)-I*sin(2/7*Pi)


Here is a diagram showing these 7 numbers.

>    UnityRootPlot(7);

[Maple Plot]


 

   3.  Complex  Roots of Unity


Roots of unity are solutions to the equation(s) :

>    z := 'z':
z^n - 1 = 0;
z^n = 0;

z^n-1 = 0

z^n = 0

 

These have a nice geometric interpretation - equally spaced spokes of a wheel - with the first spoke along the positive x axis.

>    UnityRoots(3);
UnityRootPlot(3);

1

-1/2+1/2*I*3^(1/2)

-1/2-1/2*I*3^(1/2)

[Maple Plot]

>    UnityRoots(4);
UnityRootPlot(4);

-1

1

I

-I

[Maple Plot]

>    UnityRoots(5);
UnityRootPlot(5);

1

-1/4+1/4*5^(1/2)+1/4*I*2^(1/2)*(5+5^(1/2))^(1/2)

-1/4-1/4*5^(1/2)+1/4*I*2^(1/2)*(5-5^(1/2))^(1/2)

-1/4-1/4*5^(1/2)-1/4*I*2^(1/2)*(5-5^(1/2))^(1/2)

-1/4+1/4*5^(1/2)-1/4*I*2^(1/2)*(5+5^(1/2))^(1/2)

[Maple Plot]

>    UnityRoots(6);
UnityRootPlot(6);

-1

1

-1/2*(-2+2*I*3^(1/2))^(1/2)

1/2*(-2+2*I*3^(1/2))^(1/2)

-1/2*(-2-2*I*3^(1/2))^(1/2)

1/2*(-2-2*I*3^(1/2))^(1/2)

[Maple Plot]

>    UnityRoots(13);
UnityRootPlot(13);

1

cos(2/13*Pi)+sin(2/13*Pi)*I

cos(4/13*Pi)+sin(4/13*Pi)*I

cos(6/13*Pi)+sin(6/13*Pi)*I

-cos(5/13*Pi)+sin(5/13*Pi)*I

-cos(3/13*Pi)+sin(3/13*Pi)*I

-cos(1/13*Pi)+sin(1/13*Pi)*I

-cos(1/13*Pi)-I*sin(1/13*Pi)

-cos(3/13*Pi)-I*sin(3/13*Pi)

-cos(5/13*Pi)-I*sin(5/13*Pi)

cos(6/13*Pi)-I*sin(6/13*Pi)

cos(4/13*Pi)-I*sin(4/13*Pi)

cos(2/13*Pi)-I*sin(2/13*Pi)

[Maple Plot]


There is an algebraic interpretation to these complex roots. Notice that the equation z^n -1 can be factored.

>    z^7 - 1:
% = (z-1)*simplify( %/(z-1) );

z^7-1 = (z-1)*(z^6+z^5+z^4+z^3+z^2+z+1)

>    z^13 - 1:
% = (z-1)*simplify( %/(z-1) );

z^13-1 = (z-1)*(1+z^12+z^11+z^10+z^9+z^8+z^7+z^6+z^5+z^4+z^3+z^2+z)


So an equation with this kind of polynomial can be factored - showing that there is always a root of 1. However, there are six more roots of the resuling sixth degree polynomial - these are the six other roots we saw above!

>    z^7 - 1 = 0;

z^7-1 = 0

>    (z-1)*simplify( lhs(%)/(z-1) ) = 0;

(z-1)*(z^6+z^5+z^4+z^3+z^2+z+1) = 0

>    UnityRoots(7);

1

cos(2/7*Pi)+sin(2/7*Pi)*I

-cos(3/7*Pi)+sin(3/7*Pi)*I

-cos(1/7*Pi)+sin(1/7*Pi)*I

-cos(1/7*Pi)-I*sin(1/7*Pi)

-cos(3/7*Pi)-I*sin(3/7*Pi)

cos(2/7*Pi)-I*sin(2/7*Pi)

>    UnityRootPlot(7);

[Maple Plot]

>    r6 := cos(2*Pi/7) - I*sin(2*Pi/7);

r6 := cos(2/7*Pi)-I*sin(2/7*Pi)

>    z^6+z^5+z^4+z^3+z^2+z+1;

z^6+z^5+z^4+z^3+z^2+z+1

>    subs( z= r6, %);

(cos(2/7*Pi)-I*sin(2/7*Pi))^6+(cos(2/7*Pi)-I*sin(2/7*Pi))^5+(cos(2/7*Pi)-I*sin(2/7*Pi))^4+(cos(2/7*Pi)-I*sin(2/7*Pi))^3+(cos(2/7*Pi)-I*sin(2/7*Pi))^2+cos(2/7*Pi)-I*sin(2/7*Pi)+1
(cos(2/7*Pi)-I*sin(2/7*Pi))^6+(cos(2/7*Pi)-I*sin(2/7*Pi))^5+(cos(2/7*Pi)-I*sin(2/7*Pi))^4+(cos(2/7*Pi)-I*sin(2/7*Pi))^3+(cos(2/7*Pi)-I*sin(2/7*Pi))^2+cos(2/7*Pi)-I*sin(2/7*Pi)+1

>    evalc(%): round(%);

0


So the 6 roots (excluding 1) are roots of
both  of these equations.

>    z^7 - 1 = 0;
z^6+z^5+z^4+z^3+z^2+z+1 = 0;

z^7-1 = 0

z^6+z^5+z^4+z^3+z^2+z+1 = 0



 

   4.  One Root Using DeMoivre's Theorem in Reverse


You might recall De Moivre's formula.

>    demoivre :=
z^n = (r^n)*( cos(n*theta) + I*sin(n*theta) );

demoivre := z^n = r^n*(cos(n*theta)+sin(n*theta)*I)


Now we want to use it in reverse - with some care. The opposite of an nth power is a nth root, and the opposite of multiplying by n is dividing by n.
 
      
Example  : Find a 5th root of w = 24 + 10i.


     
Step 1 . Express w in polar form

>    w := 24 + 10*I;

w := 24+10*I

>    R := abs(w);
theta := argument(w);
evalf(theta);

R := 26

theta := arctan(5/12)

.3947911197

>    w = R*'(cos(theta) + I*sin(theta))';

24+10*I = 26*cos(theta)+26*I*sin(theta)


    
Step 2 . Find the 5th root of R. (The opposite of raising
       the absolute value to the 5th power is to take the 5th root).

>    r := R^(1/5);

r := 26^(1/5)

 

    3.  Divide the argument by 5

>    phi := theta/5;

phi := 1/5*arctan(5/12)

 
    4.
Construct the root using these new values.

>    z := r*('cos'(phi) + I*'sin'(phi));

z := 26^(1/5)*(cos(1/5*arctan(5/12))+sin(1/5*arctan(5/12))*I)



Indeed, if we raise this number to the 5th power, we get back to w.

>    'z'^5 = z^5;  
 phi := evalf(theta/5):
 z := r*('cos'(phi) + I*'sin'(phi)):
 simplify(z^5);

z^5 = 26*(cos(1/5*arctan(5/12))+sin(1/5*arctan(5/12))*I)^5

24.+10.*I

   5.  Expanding One Arbitrary Root into Many


We found one root, but we know there are others. Lets see how to find them.

 
        Example 5.1 :  Find the 5th roots of -1

Lets start with a "complex" number, w = -1. We can express this number in polar form.

>    w := -1;

w := -1

>    R := abs(w);
theta := argument(w);

R := 1

theta := Pi

 
We can find a single 5th root by dividing the angle by 5. (since R = 1)

>    z1 := cos(theta/5) + I*sin(theta/5);

z1 := cos(1/5*Pi)+sin(1/5*Pi)*I


It's also apparent from the diagram that 5 times the angle of z1 is
Pi .

>    display(ComplexPlotY(z1), ComplexPlotB(w),
plottools[circle]([0,0],1, color = gold),
plottools[arc]([0,0],.8, 0..Pi/5, color = gold, thickness = 3),
plottools[arc]([0,0],.9, 0..Pi, color = gold, thickness = 3) );

[Maple Plot]


Now here is the key new idea. We need to find other angles, which when multiplied by 5, give theta plus even multiples of
Pi . We can simply solve an equation like this. We solve all of these kinds of equations in the same way.

>    theta + 3*2*Pi = 5*x;  
phi||3  := solve( %, x);

7*Pi = 5*x

phi3 := 7/5*Pi

>    for k from 1 to 5 do
  theta + (k-1)*2*Pi = 5*x;  
  phi||k  := solve( %, x);
  z||k    := cos(phi||k) + I*sin(phi||k);
  print(`-----------------------------------`);
od;

Pi = 5*x

phi1 := 1/5*Pi

z1 := cos(1/5*Pi)+sin(1/5*Pi)*I

`-----------------------------------`

3*Pi = 5*x

phi2 := 3/5*Pi

z2 := -cos(2/5*Pi)+sin(2/5*Pi)*I

`-----------------------------------`

5*Pi = 5*x

phi3 := Pi

z3 := -1

`-----------------------------------`

7*Pi = 5*x

phi4 := 7/5*Pi

z4 := -cos(2/5*Pi)-I*sin(2/5*Pi)

`-----------------------------------`

9*Pi = 5*x

phi5 := 9/5*Pi

z5 := cos(1/5*Pi)-I*sin(1/5*Pi)

`-----------------------------------`


Let's see what it looks like.

>    ComplexPlotY(z1,z2,z3,z4,z5);

[Maple Plot]




Indeed, if we raise any of these numbers to the 5th power, we get back w = -1.

>    for k from 1 to 5 do
   z||k^5 = round(evalf(z||k^5)) ;
od;

(cos(1/5*Pi)+sin(1/5*Pi)*I)^5 = -1

(-cos(2/5*Pi)+sin(2/5*Pi)*I)^5 = -1

-1 = -1

(-cos(2/5*Pi)-I*sin(2/5*Pi))^5 = -1

(cos(1/5*Pi)-I*sin(1/5*Pi))^5 = -1



    
Example 5.2 :  Find the 4th roots of i

Lets start with a "complex" number, w = i. And go through these same steps.

>    w := I;

w := I

>    R := abs(w);
theta := argument(w);

R := 1

theta := 1/2*Pi

 
We can find a single 4th root by dividing the angle by 5. (since R = 1)

>    z1 := cos(theta/4) + I*sin(theta/4);

z1 := cos(1/8*Pi)+sin(1/8*Pi)*I


It's also apparent from the diagram that 4 times the angle of z1 is
Pi/2 ,

>    display(ComplexPlotY(z1), ComplexPlotB(w),
plottools[circle]([0,0],1, color = gold),
plottools[arc]([0,0],.8, 0..Pi/8, color = gold, thickness = 3),
plottools[arc]([0,0],.9, 0..Pi/2, color = gold, thickness = 3) );

[Maple Plot]


Solve the  equations which give the same resulting angle.

>    for k from 1 to 4 do
  theta + (k-1)*2*Pi = 4*x;  
  phi||k  := solve( %, x);
  z||k    := cos(phi||k) + I*sin(phi||k);
  print(`-----------------------------------`);
od;

1/2*Pi = 4*x

phi1 := 1/8*Pi

z1 := cos(1/8*Pi)+sin(1/8*Pi)*I

`-----------------------------------`

5/2*Pi = 4*x

phi2 := 5/8*Pi

z2 := -cos(3/8*Pi)+sin(3/8*Pi)*I

`-----------------------------------`

9/2*Pi = 4*x

phi3 := 9/8*Pi

z3 := -cos(1/8*Pi)-I*sin(1/8*Pi)

`-----------------------------------`

13/2*Pi = 4*x

phi4 := 13/8*Pi

z4 := cos(3/8*Pi)-I*sin(3/8*Pi)

`-----------------------------------`


Let's see what it looks like.

>    display( ComplexPlotY(z1,z2,z3,z4), ComplexPlotB(w),
plottools[circle]([0,0],1, color = gold));

[Maple Plot]



Please take a moment to study this diagram because it illustrates the principle of complex roots. The blue square at i (0,1), is the original number. You can see the first 4th root at
Pi /8 in the first quadrant - and we know that angle is exactly 1/4 of the angle of i. Now other three roots are spread around the circle - cutting the circle into four equal pieces - but starting at the first root - not at an angle of zero.



And as before, if we raise any of these numbers to the 4th power, we get back w = i.

>    for k from 1 to 4 do
   z||k^5 = round(evalf(z||k^4)) ; od;

(cos(1/8*Pi)+sin(1/8*Pi)*I)^5 = I

(-cos(3/8*Pi)+sin(3/8*Pi)*I)^5 = I

(-cos(1/8*Pi)-I*sin(1/8*Pi))^5 = I

(cos(3/8*Pi)-I*sin(3/8*Pi))^5 = I


 

   6.  Complete Set of Complex Roots


We have seen how we can use DeMoivre's formula in reverse to find a single root of a complex number. We have also seen how a single complex root can be expanded to n distinct complex roots. Now we sew this all up, and use these ideas to find the nth roots for any complex number.

     Example  : Find ALL of the 5th roots of w = 24 + 10i.


     
1 . Express w in polar form

>    w := 24 + 10*I;

w := 24+10*I

>    R := abs(w);
theta := argument(w);
evalf(theta);

R := 26

theta := arctan(5/12)

.3947911197

>    w = R*'(cos(theta) + I*sin(theta))';

24+10*I = 26*cos(theta)+26*I*sin(theta)


    
2 . Find the 5th root of R - which is never a problem since R is always non-negative. (The opposite of raising
       the absolute value to the 5th power, is to take the 5th root).

>    r := R^(1/5);

r := 26^(1/5)

 

    3.  Divide the argument by 5 (the opposite of multiplying by 5 is dividing by 5)

>    phi := theta/5;

phi := 1/5*arctan(5/12)

 
    4.
Construct the first root using these values.

>    z||1 := r*('cos'(phi) + I*'sin'(phi));

z1 := 26^(1/5)*(cos(1/5*arctan(5/12))+sin(1/5*arctan(5/12))*I)


    5. Construct the complete set of roots by adding multiples of 2*Pi  to theta, then divide by n.

>    for k from 0 to 4 do
     z||k :=r*('cos'((theta + 2*Pi*k)/n) + I*'sin'((theta + 2*Pi*k)/n));
od;

z0 := 26^(1/5)*(cos(arctan(5/12)/n)+sin(arctan(5/12)/n)*I)

z1 := 26^(1/5)*(cos((arctan(5/12)+2*Pi)/n)+sin((arctan(5/12)+2*Pi)/n)*I)

z2 := 26^(1/5)*(cos((arctan(5/12)+4*Pi)/n)+sin((arctan(5/12)+4*Pi)/n)*I)

z3 := 26^(1/5)*(cos((arctan(5/12)+6*Pi)/n)+sin((arctan(5/12)+6*Pi)/n)*I)

z4 := 26^(1/5)*(cos((arctan(5/12)+8*Pi)/n)+sin((arctan(5/12)+8*Pi)/n)*I)

>    for k from 0 to 4 do
     z||k :=r*(    'cos'(evalf(  (theta + 2*Pi*k)/5 ) )
               + I*'sin'(evalf(  (theta + 2*Pi*k)/5 ) ) );
od;

z0 := 26^(1/5)*(cos(.7895822394e-1)+sin(.7895822394e-1)*I)

z1 := 26^(1/5)*(cos(1.335595286)+sin(1.335595286)*I)

z2 := 26^(1/5)*(cos(2.592232347)+sin(2.592232347)*I)

z3 := 26^(1/5)*(cos(3.848869409)+sin(3.848869409)*I)

z4 := 26^(1/5)*(cos(5.105506470)+sin(5.105506470)*I)

>    for k from 0 to 4 do
     z||k := evalf( r*(    'cos'( (theta + 2*Pi*k)/5 )
                       + I*'sin'( (theta + 2*Pi*k)/5 ) ));
od;

z0 := 1.912667497+.1513354546*I

z1 := .4471181897+1.865820114*I

z2 := -1.636333258+1.001804792*I

z3 := -1.458427760-1.246670703*I

z4 := .7349753320-1.772289659*I




If we raise any of these numbers to the 5th power, we get back to w!

>    z0 := 1.912667497+.1513354546*I;
'z0^5' = round(simplify(z2^5));

z0 := 1.912667497+.1513354546*I

z0^5 = 24+10*I

>    z1 := .4471181897+1.865820114*I;
'z1^5' = round(simplify(z2^5));

z1 := .4471181897+1.865820114*I

z1^5 = 24+10*I

>    z2 := -1.636333258+1.001804792*I;
'z2^5' = round(simplify(z2^5));

z2 := -1.636333258+1.001804792*I

z2^5 = 24+10*I

>    z3 := -1.458427760-1.246670703*I;
'z3^5' = round(simplify(z2^5));

z3 := -1.458427760-1.246670703*I

z3^5 = 24+10*I

>    z4 := .7349753320-1.772289659*I;
'z4^5' = round(simplify(z2^5));

z4 := .7349753320-1.772289659*I

z4^5 = 24+10*I


 

   7.  A Geometric Representation of Complex Roots


Here are some geometric representations of complex roots.

>    z := 'z':

>    w :=  3 + 3*I;   n := 3;
roots_of_unity := solve( z^3 = w, z);

w := 3+3*I

n := 3

roots_of_unity := (3+3*I)^(1/3), -1/2*(3+3*I)^(1/3)+1/2*I*3^(1/2)*(3+3*I)^(1/3), -1/2*(3+3*I)^(1/3)-1/2*I*3^(1/2)*(3+3*I)^(1/3)

>    display(  ComplexPlotB( w),
          ComplexPlotY( seq( roots_of_unity[k], k = 1..n) ),
           polarplot( abs(roots_of_unity[1]), color = gray ));

[Maple Plot]

>   

>    w :=  1 + 3*I;   n := 3;
roots_of_unity := solve( z^3 = w, z);

w := 1+3*I

n := 3

roots_of_unity := (1+3*I)^(1/3), -1/2*(1+3*I)^(1/3)+1/2*I*3^(1/2)*(1+3*I)^(1/3), -1/2*(1+3*I)^(1/3)-1/2*I*3^(1/2)*(1+3*I)^(1/3)

>    display(  ComplexPlotB( w),
          ComplexPlotY( seq( roots_of_unity[k], k = 1..n) ),
           polarplot( abs(roots_of_unity[1]), color = gray ));

[Maple Plot]

>    display(  ComplexPlotB( w),
          ComplexPlotY( seq( roots_of_unity[k], k = 1..n) ),
          polarplot( evalf( abs( roots_of_unity[1])), color = gray ));

[Maple Plot]


 

   8. General Formula


Here is the formula for the powers of complex number in polar form.

>    reset();
z^n = (r^n)*( cos(n*theta) + I*sin(n*theta) );

a, b, z, w, R, n, k, theta, phi

z^n = (26^(1/5))^n*(cos(n*theta)+sin(n*theta)*I)



Here is the formula for the powers of complex number in polar form.

>    w = R*('cos'(theta) + I*'sin'(theta));

w = R*(cos(theta)+sin(theta)*I)

>    z = (R^(1/n))*( cos((theta + 2*Pi*k)/n) + I*sin((theta + 2*Pi*k)/n));

z = R^(1/n)*(cos((theta+2*Pi*k)/n)+sin((theta+2*Pi*k)/n)*I)


For example, for n = 7, we have ....

>    for k from 0 to 6 do  
z||k = (R^(1/n))*( cos((theta + 2*Pi*k)/n) + I*sin((theta + 2*Pi*k)/n)); od;

1.912667497+.1513354546*I = R^(1/n)*(cos(theta/n)+sin(theta/n)*I)

.4471181897+1.865820114*I = R^(1/n)*(cos((theta+2*Pi)/n)+sin((theta+2*Pi)/n)*I)

-1.636333258+1.001804792*I = R^(1/n)*(cos((theta+4*Pi)/n)+sin((theta+4*Pi)/n)*I)

-1.458427760-1.246670703*I = R^(1/n)*(cos((theta+6*Pi)/n)+sin((theta+6*Pi)/n)*I)

.7349753320-1.772289659*I = R^(1/n)*(cos((theta+8*Pi)/n)+sin((theta+8*Pi)/n)*I)

cos(1/5*Pi)-I*sin(1/5*Pi) = R^(1/n)*(cos((theta+10*Pi)/n)+sin((theta+10*Pi)/n)*I)

z6 = R^(1/n)*(cos((theta+12*Pi)/n)+sin((theta+12*Pi)/n)*I)

>   


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