Complex Number!

[B. Wong]

As I found out early, in US the maths standard should be higher
than in Australia. I found one question on complex number
that even my teacher could do it (at least I disagree with her
application)! If you can help me I will be very happy!

Q: Solve X^5 = 1 by De Moivre's theorem and indicate roots on
a circle of radius 1 in an Argand diagram. Express the roots
in the mod-arg form.
Find the area of the regular pentagon formed by the five points
representing these roots.
Solve the equation X^4 + X^3 + X^2 + X + 1 = 0.
Deduce that cos2PI/5 and cos4PI/5 are the roots of
4X^2 + 2X - 1 = 0.

(the problem for me lies on how to deduce the roots beside
using subsitution)

Please Email to me and I'll be very appreciated!

From Billy Wong, year 12, Australia.

[Steven T Spallone]

B. Wong (bre...@real.net.au) wrote:
: As I found out early, in US the maths standard should be higher

: than in Australia. I found one question on complex number
: that even my teacher could do it (at least I disagree with her
: application)! If you can help me I will be very happy!

: Q: Solve X^5 = 1 by De Moivre's theorem and indicate roots on
: a circle of radius 1 in an Argand diagram. Express the roots
: in the mod-arg form.

We use the fact that e^(ix)= cos x + i*sin x.

We see 1 = cos 0 + i * sin 0 = e^0i. The trick is that sine and cosine
are periodic, with period 2pi. So 1= cos 2pi + i*sin 2pi = e^(2pi*i).

So what's a fifth root of e^(2pi*i)? Why, e^((2pi/5)*i), of course.
Check that e^(n(2pi/5)*i), for n=1,2,3,4,5, are all the roots.

And these equal cos(2pi*n/5) + i sin(2pi*n/5), by our important fact. I
forget what mod-arg form is.

: Find the area of the regular pentagon formed by the five points
: representing these roots.

These roots all fall on the unit circle. Draw a picture for clarity.
Break the pentagon into the 5 natural isosceles triangles, each with a
point in the center and so forth. Look at the first one, the one above
the positive real axis. The central angle is (2pi/5), and the congruent
sides are equal to 1. The answer follows from some straightforward trig.

: Solve the equation X^4 + X^3 + X^2 + X + 1 = 0.

Check that this polynomial times X-1 is X^5-1. We already solved it.

: Deduce that cos2PI/5 and cos4PI/5 are the roots of

: 4X^2 + 2X - 1 = 0.

cos 2pi/5 is the average of X and X^4 where X = the root where n=1. Use
the equations X solves and this will pop out.

: (the problem for me lies on how to deduce the roots beside
: using subsitution)

: Please Email to me and I'll be very appreciated!

: From Billy Wong, year 12, Australia.

Cheers,

-Steven, year 19 going on 20, Philly

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[Tony2back]

In article <31893D...@real.net.au>, "B. Wong" <bre...@real.net.au>
writes:

>
>Q: Solve X^5 = 1 by De Moivre's theorem and indicate roots on
> a circle of radius 1 in an Argand diagram. Express the roots
> in the mod-arg form.

> Find the area of the regular pentagon formed by the five points
> representing these roots.

> Solve the equation X^4 + X^3 + X^2 + X + 1 = 0.

> Deduce that cos2PI/5 and cos4PI/5 are the roots of
> 4X^2 + 2X - 1 = 0.
>

> (the problem for me lies on how to deduce the roots beside
> using subsitution)
>
>

Write the equation in the form z^5 = cos(2k.pi) + i.sin(2k.pi) where k =
0, 1,...

Take 5th root of each side z = [cos(2k.pi) + i.sin(2k.pi)]^(1/5)

using DeMoivre's theorem z = [cos(2k.pi/5) + i.sin(2k.pi/5)] k = 0,
1,..., 4

and so z1 = cos(0) + i.sin(0)
z2 = cos(2.pi/5) + i.sin(2.pi/5)
z3 = cos(4.pi/5) + i.sin(4.pi/5)
z4 = cos(6.pi/5) + i.sin(6.pi/5)
z5 = cos(8.pi/5) + i.sin(8.pi/5)

These points can be plotted on an Argand diagram as 5 points equally
spaced round the unit circle, starting at the point (1,0) on the real
axis.

The area of the pentagon formed by these points is 5 times the area of a
triangle formed by two adjacent 'spokes' from the origin to the points.

Area of pentagon = 5 * (1/2)(1)(1)sin(2.pi/5) = (5/2)sin(2.pi/5)+

Solve x^4 + x^3 + x^2 + x + 1 = 0

It is clear that if x = cos(2.pi/5) + i.sin(2.pi/5) then
x^2 = cos(4.pi/5) + i.sin(4.pi/5)
x^3 = cos(6.pi/5) + i.sin(6.pi/5)
x^4 = cos(8.pi/5) + i.sin(8.pi/5)

and these are four of the roots of x^5 = 1 . If we add 1as the remaining
root, then the sum x^4 + x^3 + x^2 + x + 1 is the sum of the roots of x^5
- 1 = 0. Since the coefficient of x^4 is zero in this last equation, the
sum of the roots is zero, and that is what we require.

Hence x = cos(2.pi/5) + i.sin(2.pi/5) is a solution of the quartic
equation. It is also true that x = cos(4.pi/5) + i.sin(4.pi/5) is a
solution, since powers of this root will generate all the positions round
the pentagon. For an equation with real coefficients, another solution of
the quartic must be the conjugate complex root. So x = cos(2.pi/5) -
i.sin(2.pi/5) is also a root of the quartic. It follows that [x -
cos(2.pi/5) - i.sin(2.pi/5)][x - cos(2.pi/5) + i.sin(2.pi/5)] will be a
factor of the quartic.

This can be written [x - cos(2.pi/5)]^2 + [sin(2.pi/5)]^2

= x^2 - 2xcos(2.pi/5) + cos^2(2.pi/5) +
sin^2(2.pi/5)

= x^2 - 2xcos(2.pi/5) + 1

If you divide this into the expression x^4 + x^3 + x^2 + x + 1 you obtain
a remainder

2xcos(2.pi/5)[4cos^2(2.pi/5) + 2cos(2.pi/5) - 1] - [4cos^2(2.pi/5) +
2cos(2.pi/5) - 1]

and this remainder must be zero. This requires that

4cos^2(2.pi/5) + 2cos(2.pi/5) - 1 = 0 and so we can say that

cos(2.pi/5) must be a root of 4x^2 + 2x -1 = 0

also cos(4.pi/5) must be a root, and so these are the roots of the
quadratic.

Anthony Hugh Back