IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => IB => Math => Topic started by: Matricaria on October 17, 2011, 11:00:33 pm

Title: Complex Numbers (De Moivre's)
Post by: Matricaria on October 17, 2011, 11:00:33 pm
- Use de moivre's theorem to obtain solutions for z^3-1=0
- 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.
- Repeat the above for the quations z^4-1=0 and z^5-1=0. Comment on you results and try to formulate a conjecture.

So I worked out the roots for z^3, z^4, and z^5 using de Moivre's
for z^3: 1, -1/2 + i*sqrt(3/2), -1/2 - i*sqrt(3/2)

for z^4: 1, -1, i, and -i

for z^5: 1, cos(2pi/5) + isin(2pi/5), cos(4pi/5) + isin(4pi/5), cos(6pi/5) + isin(6pi/5), and cos(8pi/5) + isin(8pi/5)

I also plotted the roots for all three equations on an argand diagram and a unit circle and my observation was that any root for z^n=1 will lie an the unit circle and that all roots of z^n=1 are equally spaced around the circle..

Then I calculated the distance between each two roots to find out the following:
For z^3: sqrt3

For z^4: sqrt2

For z^5: Approx value 1.175570505

Then I tried to formulate a general equation for the distance between any two neighboring roots of z^n=1, and I came up with this:
|cis2pi/n - 1|

NOW: My findings are:

z^n=1 has n roots given by z=cis(k 2pi/n), where k is the number 0, 1, ..., n-1.

The distance between 2 neighboring roots is |cis2pi/n - 1|.

However, I'm afraid this is not a conjecture, but just fact...


Can you please tell me how to put these findings into a conjecture?
Title: Re: Complex Numbers (De Moivre's)
Post by: astarmathsandphysics on October 19, 2011, 08:41:42 am
here