- Use graphing software to plot these roots on an Argand diagram as well as a unit circle with centre origin.
- Choose a root and draw line segments from this root to the other two roots.
- Measure these line segments and comment on your results.
As we see that the length of line segments are equal to each other. (one segment is 0.01 unit bigger than the others because of accuracy of graphing software)
Repeat the above for the equations
and
- Comment on your results and try to formulate a conjecture.
For
0
For
Our conjecture is finding distance with using argument difference . In analytic geometry, the distance between two points can be found with distance formula ;
As we seen on the graph and second part our ‘r’ value is equal to 1. When we remember the ‘r’ formula ;
From there r=1 so
=1 , lets check it with trigonometry ;
a is our x(cos) value and b is our y(sin) value .
lets say our argument is α ;
Lets connect them to each other ;
Lets use
to see our conjecture ; our first argument is cis(0) the other one is cis(72) ;
value cos(0) ,
is cos(72)
value is sin(0)
value is sin(72).
Now we will put them on equation .
1+1-2(cos72*cos 0 +sin72*sin0)
We will use this trigonometric identities ;
and
When we put our value at these formula ;
=cos 72
Lastly our final formula is ;
Factorize
-1= (z-1)(
-1= (z-1)(
-1= (z-1)(
-
Use graphing software to test your conjecture for some more values of n € Z+ and make modifications to your conjecture if necessary.
Lets try it for n=6
Testing for Aw distance ;
a=cis0 and w is cis120
=1.732050… ≅17.3 (because of accuracy of software)
or testing for Az distance ;
=1= 1 from graph .
PART B
Use de Moivre’s theorem to obtain solutions to
For
=i
where where k = 0, 1, 2
We found ; cis(30),cis(150),cis(270)
For
=i
where where k = 0, 1, 2,3
We found ;
Cis(22.5) cis (112.5) cis(202.5) cis(292.5)
For
=i
where where k = 0, 1, 2,3,4
We found ;
Cis(18) cis (90) cis(162) cis(234) cis(306)
- Use graphing software to represent each of these solutions on an Argand diagram.
For
=i
For
=i
For
=i
Generalize and prove your results for i
-
=a + bi , where |a +bi |=1.
|a +bi |=1 so it means that ;
+
1=
As we can see r=1 so
=a+bi= r cis(θ)
When we use De Moivre’s theorem, it states ;
For our solution we will use reverse way of it ;
r=1 so we dont need r at equation ;
z=
let’s find the
value with tangent ;
tan(
=
=
and
=
so z=cis(
)
-
What happens when |a +bi |≠1?
That doesn’t apply when when |a +bi |≠1 because r value should equal to 1.
Resources
http://en.wikipedia.org/wiki/Imaginary_number