# 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?