Cosine and sine of 2Pi/5 problem

[cdummie]

We have a complex number ω = cos(2π/5) +isin(2π/5) and we have two complex numbers a and b, such that:

a= ω + ω⁴ and b= ω² + ω³. I have to prove that a + b = -1 and a*b= -1. Then, based on that, determine cos(2π/5) and sin(2π/5).

I've tried to solve this using trigonometry. First a+b, i got:

cos(2Pi/5) + cos(4Pi/5) + cos(6Pi/5) + cos(8Pi/5) + i[ sin(2Pi/5) + sin(4Pi/5) + sin(6Pi/5) + sin(8Pi/5)]

then i tried expressing cos(6Pi/5) as cos(2Pi/5 + 5Pi/5) = cos(2Pi/5)cos(4Pi/5) + sin(2Pi/5)sin(4Pi/5)

and cos(4Pi/5) as cos²(2Pi/5) - sin²(2Pi/5). But that wasn't helpful. Knowing that a + b equals -1 that means that this whole expression should end up as cosPi + isinPi since that is equal to -1. But i just can't get there.

 

[BvU]

Are you familiar with the Euler formula ?
Mark your $\omega^n$ terms in a and b on a unit circle in the complex plane and see what that brings you...

 

[cdummie]

Are you familiar with the Euler formula ?
Mark your $\omega^n$ terms in a and b on a unit circle in the complex plane and see what that brings you...

I know Euler's formula in this example i'd have: $ω=e^\frac{i2π}{5}$ or for a + b i would have: $a+b = e^\frac{i2π}{5} + e^\frac{i4π}{5} + e^\frac{i6π}{5} + e^\frac{i8π}{5}$ but i don't know how this helps...

When i draw a unit circle i don't see much, i mean i see the points that represent corresponding complex numbers but i don't see how it gets me closer to the solution.

 

[BvU]

When i draw a unit circle i don't see much, i mean i see the points that represent corresponding complex numbers but i don't see how it gets me closer to the solution.

Do you see the same thing I see ? Addition of complex numbers is like vector addition in the x,y plane.
Can you post your sketch ?

 

[cdummie]

Do you see the same thing I see ? Addition of complex numbers is like vector addition in the x,y plane.
Can you post your sjetch ?

Here is the sketch:

But, i can't see what will i get exactly by just looking at this picture (i'd like you to explain it to me though), and i'd like to know if there's any algebraic way to solve this?

 

[BvU]

Addition of complex numbers is like vector addition in the x,y plane.

So where are a and b in the picture ?

 

[cdummie]

So where are a and b in the picture ?

a would be somewhere in the middle of the positive part of x-axis in the circle (even though it seems a little bit closer to zero, but this is just a sketch) and b is on the negative part of the x-axis in the circle little bit closer to (-1,0)

 

[Samy_A]

Are you sure that you have written the problem correctly? From your problem description and your graph, it seems to me that a + b should equal 0. ω² + ω³ add to -2cos(36°), while ω + ω⁴ add to +2cos(72°). It seemed simpler to write these angles in degrees...
I do get a*b = -1, but I don't get a + b = -1.

I'm confused now. I do get a+b=-1 with the same results for a and b as you.
Mark44 edit: I made a mistake in my calculations, so the text that Samy_A is quoting is in error. I have also deleted my reply that is quoted above.

 

[Mark44]

I'm confused now. I do get a+b=-1 with the same results for a and b as you.

Oops, my mistake. I'll go back and edit my earlier post.

 

[fresh_42]

Have you calculated $ω^5, ω^6, ω^7$ to calculate $a \cdot b$? And are you familiar with polynomial divisions?
There is an interesting generalization of the 3rd binomial formula $(x-1)(x+1) = x^2 -1$ for $x^n -1$.

 

[BvU]

a would be somewhere in the middle of the positive part of x-axis in the circle (even though it seems a little bit closer to zero, but this is just a sketch) and b is on the negative part of the x-axis in the circle little bit closer to (-1,0)

Check b: it should be outside the circle (a.b is supposed to give -1, remember ?)

And: what is $\omega^5$ ?

 

[WWGD]

Maybe you can use the fact that if $w$ is an n-th root of unity, then EDIT $1+ w+w^2+...+w^n =0$ , changing the proof a bit to adapt it to your situation.

 

[cdummie]

Check b: it should be outside the sircle (a.b is supposed to give -1, remember ?)

And: what is $\omega^5$ ?

Oh i see now, b IS outside, but it's somewhere outside of circle, just like a is somewhere inside circle, that won't help if i don't know where they are exactly.
$\omega^5 = 1$

 

[cdummie]

Check b: it should be outside the circle (a.b is supposed to give -1, remember ?)

And: what is $\omega^5$ ?

So, what can i do next?

 

[Mark44]

We have a complex number ω = cos(2π/5) +isin(2π/5) and we have two complex numbers a and b, such that:

a= ω + ω⁴ and b= ω² + ω³. I have to prove that a + b = -1 and a*b= -1. Then, based on that, determine cos(2π/5) and sin(2π/5).

Here is the sketch:
View attachment 92127

Use your drawing.
a + b = ω + ω⁴ + ω² + ω³
ω and ω⁴ are the two complex numbers on the right side of the drawing. They have the same x-coordinate and their y-coordinates are opposites. Similarly, ω² and ω³ are on the left side of the drawing. Their x-coordinates are equal and their y-coordinates are opposite in value.

Ordinary trig can be used to show that a + b = -1.

For the product ab, it's helpful to know that ω, ω², ω³,ω⁴, and ω⁵ are all equally spaced around the unit circle, and that higher powers simply retrace certain lower powers. By that I mean that ω⁶ = ω, ω⁷ = ω², and so on.

 

[BvU]

It's difficult to help you without giving away too much. So far we've established that a and b are real and that 0<a<1 and -2<b<-1.

You have also seen that $\omega^5 = 1$ which makes ab = a + b $\ \ \$ (clear ?)

Unfortunately, Mark's 'ordinary trig' doesn't help me other than that I can confirm a + b = -1 numerically. But the exercised wanted us to prove it.

For that we can now do something useful with $0 = \omega^5 - 1\ \ \$ -- see hint by fresh in post #10

lovely exercise !
--

 

[Mark44]

Unfortunately, Mark's 'ordinary trig' doesn't help me other than that I can confirm a + b = -1 numerically.

Yes, you're correct. I just used a calculator to get a value, but didn't show it algebraically.

lovely exercise !

Yes!

To expand slightly on the hint from fresh_42 that BvU mentions, the idea is related to these facts:
$(x - 1)(x + 1) = x^2 - 1$
$(x - 1)(x^2 + x + 1) = x^3 - 1$
$(x - 1)(x^3 + + x^2 + x + 1) = x^4 - 1$
and so on. This idea can be helpful in evaluating a*b.

 

[ehild]

a + b = ω + ω⁴ + ω² + ω³.
You can add 1 to it and rearrange the terms: a+b+1= 1+ω+ω²+ω³+ω⁴. It is a the sum of the first five elements of a geometric progression. You certainly know the formula of the sum.
If you expand ab+1 and use that ω⁵ = 1, you get the same sum.

 

[BvU]

a + b = ω + ω⁴ + ω² + ω³.
...

Didn't we want to leave something for cd to discover ?

 

[fresh_42]

Didn't we want to leave something for cd to discover ?

How about the stereographic projection?

 

[BvU]

Stereographic projection ? If you mean the picture: made us see what we were after !

 

[fresh_42]

Stereographic projeetion ? If you mean the picture: made us see what we were after !

I mean it would be something about complex numbers to explore and possibly a natural next eureka moment. It would fit the context.

 

[BvU]

Not so sure. I haven't figured out how to get ab = -1 out of the picture, nor a+b = -1

 

[fresh_42]

Not so sure. I haven't figured out how to get ab = -1 out of the picture, nor a+b = -1

Neither did I but after a short glimpse I haven't even tried. It was much harder not to post the solution right away. Therefore my post #10. My suggestion about stereographic projection was merely meant as an answer to your post about something left to discover. I remember a calculation on complex numbers which took me two substitutions and a couple of tricks while the author just called it obvious. It just seemed to me to be another useful tool when dealing with complex numbers. We already had Euler and roots of unity.

 

[ehild]

Didn't we want to leave something for cd to discover ?

That was discovered by the OP already (Post #3) and also appeared in Mark's Post #15, adding 1 appeared inWWGD's Post #12 , only nobody mentioned geometric progression , which the OP is certainly familiar with, and provides an easy solution.

 

[ehild]

Something else to discover about 72°, but not connected to the OP.

Halve the 72° angle of the isoceles triangle in the picture. The blue triangle is similar to the original one, and both the pink and blue triangles are isosceles. From similarity, you get the relation between a and b, and it is easy to derive the sine and cosine of 72° from here.