# Homework Help: 5 degree equation contains imaginary value?

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1. Jun 20, 2017

### Helly123

1. The problem statement, all variables and given/known data

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

3. 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?

2. Jun 20, 2017

### Buffu

Are you allowed to use polar form ?

3. Jun 20, 2017

### Staff: Mentor

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 $\sum_{k=0}^2 \binom 2 k a^{2-k} b^k$.

4. Jun 20, 2017

### Helly123

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

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

5. Jun 20, 2017

### Staff: Mentor

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.

6. Jun 20, 2017

### Helly123

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

7. Jun 20, 2017

### Staff: Mentor

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

8. Jun 20, 2017

### 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?

9. Jun 20, 2017

### Staff: Mentor

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 $\sin(5\pi/3)$.

10. Jun 20, 2017

### mjc123

z = re = r(cosθ + i*sinθ). In this case r = 1 and θ = π/3
z5 = r5ei5θ = ei5π/3 = e-iπ/3 = 1/2 - i√3/2

11. Jun 20, 2017

### Buffu

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

Last edited: Jun 20, 2017
12. Jun 20, 2017

### Helly123

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

13. Jun 20, 2017

### Buffu

14. Jun 20, 2017

### Helly123

15. Jun 20, 2017

### Buffu

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

16. Jun 20, 2017

### Helly123

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

17. Jun 20, 2017

### Helly123

How can ei5π/3 = e-iπ/3 ? @mjc123

18. Jun 20, 2017

### Buffu

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

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

19. Jun 21, 2017

### mjc123

Adding a multiple of 2π to the angle brings you back to the same point. So e = ei(θ+2nπ)
So ei5π/3 = ei(5π/3 -2π) = e-iπ/3. Draw the Argand diagram to see this.

20. Jun 21, 2017

### Helly123

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

21. Jun 21, 2017

### Staff: Mentor

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.

22. Jun 23, 2017

### 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 ??

23. Jun 23, 2017

### Ray Vickson

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

Last edited: Jun 23, 2017
24. Jun 23, 2017

### mjc123

You've missed out the i and written phi for pi, but
ei5π/3 = ei*300° = ei*(300°-360°) = ei(-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.

25. Jul 2, 2017

### Helly123

Why 300-360, not 360-300 ?