5 degree equation contains imaginary value?

[Helly123]

Homework Statement

Homework Equations

polynomial equation $$(a+b)^2 = \sum_{k=0}^{n} \binom nk a^{n-k} b^k $$

The Attempt at a Solution

i get $$ \frac {1} {32} + \frac {5} {16}\frac {3^{0.5}i} {2} +\frac {10} {8}\frac {3i^2} {4} + \frac {10} {4}\frac {3^{1.5}i^3} {8} + \frac {5} {2} \frac {9 i^4} {16} + \frac {9.3^{0.5}i^5} {32} $$

and it didn't work out. can anyone show me other way to solve it?

 

[Buffu]

Homework Statement

View attachment 205810

Homework Equations

polynomial equation $$(a+b)^2 = \sum_{k=0}^{n} \binom nk a^{n-k} b^k $$

The Attempt at a Solution

i get $$ \frac {1} {32} + \frac {5} {16}\frac {3^{0.5}i} {2} +\frac {10} {8}\frac {3i^2} {4} + \frac {10} {4}\frac {3^{1.5}i^3} {8} + \frac {5} {2} \frac {9 i^4} {16} + \frac {9.3^{0.5}i^5} {32} $$

and it didn't work out. can anyone show me other way to solve it?

Are you allowed to use polar form ?

 

[Mark44]

Homework Statement

View attachment 205810

Homework Equations

polynomial equation $$(a+b)^2 = \sum_{k=0}^{n} \binom nk a^{n-k} b^k $$

The Attempt at a Solution

i get $$ \frac {1} {32} + \frac {5} {16}\frac {3^{0.5}i} {2} +\frac {10} {8}\frac {3i^2} {4} + \frac {10} {4}\frac {3^{1.5}i^3} {8} + \frac {5} {2} \frac {9 i^4} {16} + \frac {9.3^{0.5}i^5} {32} $$

and it didn't work out. can anyone show me other way to solve it?

I didn't verify your work above, but you should replace i^2 by -1, and i^3 by -i, and so on.

This problem is much easier if you use @Buffu's suggestion, assuming you know about polar form for complex numbers.

BTW, your "relevant equation" isn't relevant, as it's wrong. With an exponent of 2 on the left side, the summation should be $\sum_{k=0}^2 \binom 2 k a^{2-k} b^k$.

 

[Helly123]

I didn't verify your work above, but you should replace i^2 by -1, and i^3 by -i, and so on.

This problem is much easier if you use @Buffu's suggestion, assuming you know about polar form for complex numbers.

BTW, your "relevant equation" isn't relevant, as it's wrong. With an exponent of 2 on the left side, the summation should be $\sum_{k=0}^2 \binom 2 k a^{2-k} b^k$.

why should I replace that? i^2 = -1 ? i^3 = -i? why?

Are you allowed to use polar form ?

yes, what exactly the purpose of using polar form?
w = 1/2 + ( (root3) / 2) i
r = 1, tetha = 60degrees??

 

[Mark44]

why should I replace that? i^2 = -1 ? i^3 = -i? why?

Because by doing this you can reduce the 6 terms you showed to only 2 terms.

yes, what exactly the purpose of using polar form?
w = 1/2 + ( (root3) / 2) i
r = 1, tetha = 60degrees??

Multiplication of complex numbers is much easier in polar form. If $z_1 = r_1e^{i\theta_1}$ and $z_2 = r_2e^{i\theta_2}$, then $z_1z_2= r_1r_2e^{i(\theta_1 + \theta_2)}$

By the way, this Greek letter -- $\theta$ -- is theta, not tetha.

 

[Helly123]

Because by doing this you can reduce the 6 terms you showed to only 2 terms.

Multiplication of complex numbers is much easier in polar form. If $z_1 = r_1e^{i\theta_1}$ and $z_2 = r_2e^{i\theta_2}$, then $z_1z_2= r_1r_2e^{i(\theta_1 + \theta_2)}$

By the way, this Greek letter -- $\theta$ -- is theta, not tetha.

i see..
replacing that not changing the value, so i^2 equal to -1 ?
i^3 equal to -i? what theory is used?
so i^4 = (i^2)^2 = (-1)^2 = 1?
i^5 = i^4 . i = i

 

[Mark44]

i see..
replacing that not changing the value, so i^2 equal to -1 ?
i^3 equal to -i? what theory is used?
so i^4 = (i^2)^2 = (-1)^2 = 1?
i^5 = i^4 . i = i

Yes to all. $i^2 = -1$ by definition. $i^3 = i^2 i = -1 i = -i$, and so on.

 

[Helly123]

I got the answer, by using conventional method. I would like to know how polar form can solve this problem... @Buffu
@Mark44 do you know how too?

 

[Mark44]

I got the answer, by using conventional method. I would like to know how polar form can solve this problem... @Buffu
@Mark44 do you know how too?

Leaving the complex number in Cartesian (or rectangular) form will work, but it takes more work.

In polar form $\omega = \frac 1 2 + \frac {\sqrt 3} 2 i = e^{i \pi/3}$. Note that because $|\omega| = 1$ the r factor in the polar form is just 1, and I don't show it.
$\omega^5 = 1^5 e^{i5 \pi/3}$, which represents a complex number whose angle is $5\pi/3$ and whose magnitude is (still) 1.This exponential polar form can be converted into trig polar form, because $e^{i\theta} = \cos(\theta) + i\sin(\theta)$, due to Euler's Formula.

So all you have to do for this problem is to evaluate $\cos(5\pi/3)$ and $\sin(5\pi/3)$.

 

[mjc123]

z = re^iθ = r(cosθ + i*sinθ). In this case r = 1 and θ = π/3
z⁵ = r⁵e^i5θ = e^i5π/3 = e^-iπ/3 = 1/2 - i√3/2

 

[Buffu]

I got the answer, by using conventional method. I would like to know how polar form can solve this problem... @Buffu
@Mark44 do you know how too?

I think you should have used $\omega^3 = -1$, then you only need to evaluate $\omega^2$ which is easier .

 

[Helly123]

I think you should have used $\omega^3 = 1$, then you only need to evaluate $\omega^2$ which is easier .

How w = 1? Isnt w^2 = -1 by definition?

 

[Buffu]

How w = 1? Isnt w^2 = -1 by definition?

I edited my previous post. It should be $\omega^3 = -1$

$-\omega$ is a cube root of unity.

See here http://www.math-only-math.com/the-cube-roots-of-unity.html.

 

[Helly123]

I edited my previous post. It should be $\omega^3 = -1$

$-\omega$ is a cube root of unity.

See here http://www.math-only-math.com/the-cube-roots-of-unity.html.

You meant w^3 = 1?

 

[Buffu]

You meant w^3 = 1?

No I mean $\omega^3 = -1 \implies (-\omega)^3 = 1$. Look at the link in my previous you can see three cube roots of $1$.

 

[Helly123]

No I mean $\omega^3 = -1 \implies (-\omega)^3 = 1$. Look at the link in my previous you can see three cube roots of $1$.

Hmm on the link i saw the property is w^3 = 1
W^5 = w^2
W^5 = 1 + root3 i + 3/4 i^2
While i^2 = -1
= 1 + root3 i - 3/4
= -1/4 + 4root 3 i / 4

It didnt work out.
If w^3 = -1
Then w^5 = 1/4 - 4root 3 i/4

 

[Helly123]

z = re^iθ = r(cosθ + i*sinθ). In this case r = 1 and θ = π/3
z⁵ = r⁵e^i5θ = e^i5π/3 = e^-iπ/3 = 1/2 - i√3/2

How can e^i5π/3 = e^-iπ/3 ? @mjc123

 

[Buffu]

Hmm on the link i saw the property is w^3 = 1
W^5 = w^2
W^5 = 1 + root3 i + 3/4 i^2
While i^2 = -1
= 1 + root3 i - 3/4
= -1/4 + 4root 3 i / 4

It didnt work out.
If w^3 = -1
Then w^5 = 1/4 - 4root 3 i/4

In that page $\omega^\prime = \dfrac{- 1 \pm \sqrt{3} i}{2}, 1$ whereas here $\omega =\dfrac{+1 + \sqrt{3} i}{2} = -\dfrac{1 + \sqrt{3} i}{2} = -\omega^\prime$.

How can e^i5π/3 = e^-iπ/3 ? @mjc123

$\exp(5i\pi/3) = \cos 5\pi/3 + i \sin 5\pi/3 = 1/2 + -\sqrt{3}i/2 = \exp(-\pi i/3)$

 

[mjc123]

How can ei5π/3 = e-iπ/3 ? @mjc123

Adding a multiple of 2π to the angle brings you back to the same point. So e^iθ = e^i(θ+2nπ)
So e^i5π/3 = e^{i(5π/3 -2π)} = e^-iπ/3. Draw the Argand diagram to see this.

 

[Helly123]

Adding a multiple of 2π to the angle brings you back to the same point. So e^iθ = e^i(θ+2nπ)
So e^i5π/3 = e^{i(5π/3 -2π)} = e^-iπ/3. Draw the Argand diagram to see this.

why do I have to do it? If it's still theta?

 

[Mark44]

How can e^i5π/3 = e^-iπ/3 ?

why do I have to do it? If it's still theta?

You're missing the point -- it's not "still theta." $-\pi/3$ and $5\pi/3$ are different angles but $e^{-i\pi/3}$ and $e^{i5\pi/3}$ represent the same complex numbers, with terminal points at the same location on the unit circle. Drawing the Argand diagram, as @mjc123 suggests, makes this clear. So no, you don't have to do this, but doing so can help you understand it better.

To my way of thinking, any time you can combine both an algebraic explanation with drawing, it makes things more understandable.

 

[Helly123]

how if my theory is like this
$$ e^{5phi/3} = e^{300degrees} = e^{360-300} = e^{60} = cos 60 + i sin 60 = 1/2 - i root3/2 ??

 

[Ray Vickson]

how if my theory is like this
$$ e^{5phi/3} = e^{300degrees} = e^{360-300} = e^{60} = cos 60 + i sin 60 = 1/2 - i root3/2 ??

You cannot possibly mean what you wrote: $e^{60}$, which is about $1.142 \times 10^{26}$, and this is not even close to $1/2 - i \sqrt{3}/2$.

 

[mjc123]

You've missed out the i and written phi for pi, but
e^i5π/3 = e^i*300° = e^{i*(300°-360°)} = e^i(-60°) = cos(-60°) + i sin (-60°) = cos 60° - i sin 60°C = 1/2 - i√3/2
It's 300 - 360, not 360 - 300. You can add (or subtract) any whole multiple of 360° to your original angle, and get the same angle.

 

[Helly123]

You've missed out the i and written phi for pi, but
e^i5π/3 = e^i*300° = e^{i*(300°-360°)} = e^i(-60°) = cos(-60°) + i sin (-60°) = cos 60° - i sin 60°C = 1/2 - i√3/2
It's 300 - 360, not 360 - 300. You can add (or subtract) any whole multiple of 360° to your original angle, and get the same angle.

Why 300-360, not 360-300 ?

 

[Mark44]

Why 300-360, not 360-300 ?

Because $\frac {5\pi} 3$ (same measure as 300°) is an angle in the 4th quadrant. 60° is in the 1st quadrant.

 

[mjc123]

Why 300-360, not 360-300 ?

What did I write immediately after that?
Did you do what I suggested and draw the Argand diagram? That should make it clear. Adding or subtracting any integer multiple of 2π (360°) takes you round a full circle and comes back to the same point. Subtracting the angle from 2π is quite different!

 

[Helly123]

What did I write immediately after that?
Did you do what I suggested and draw the Argand diagram? That should make it clear. Adding or subtracting any integer multiple of 2π (360°) takes you round a full circle and comes back to the same point. Subtracting the angle from 2π is quite different!

i try to do what you suggested before, i just get it. So -60 has same 'appearance ' as 300 degrees? So cos 300 = cos -60

 

[Helly123]

What did I write immediately after that?
Did you do what I suggested and draw the Argand diagram? That should make it clear. Adding or subtracting any integer multiple of 2π (360°) takes you round a full circle and comes back to the same point. Subtracting the angle from 2π is quite different!

so if the question find cos -60. Then cos -60= cos (360-300) ?

 

[mjc123]

So -60 has same 'appearance ' as 300 degrees? So cos 300 = cos -60

Yes.

so if the question find cos -60. Then cos -60= cos (360-300) ?

In this case it happens to be, but be careful. Sin -60° is not equal to sin (360°-300°). -60° is equivalent to 360°-60° = 300°, not to 360°-300° = 60°.
cos -θ = cos 2π-θ = cosθ. sin -θ = sin 2π-θ = -sinθ
Distinguish between making a full circle i.e. θ → 2nπ+θ, and reflecting about the x axis, i.e. θ → -θ = 2π-θ. They are not the same operation, and the latter does not give the same angle.