Where would I start in finding the real and complex solutions to the equation x^3+1=0?

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- Mar 31st 2007, 09:02 PMgretchenReal and Complex solutions
Where would I start in finding the real and complex solutions to the equation x^3+1=0?

- Mar 31st 2007, 09:34 PMJhevon
let's solve this normally

x^3 + 1 = 0

=> x^3 = -1

=> x = (-1)^(1/3)

the real solution is of course, -1. since (-1)^3 = -1

there are no complex solutions to this as far as i can tell, unless we want to make things complicated and do something like i^(2/3), that way, (i^(2/3))^3 = i^2 = -1, and we could try to come up with a formula for all fractional powers of i that causes this to happen, but that's just overkill as far as i'm concerned - Mar 31st 2007, 10:22 PMSoroban
Hello, Gretchen!

Quote:

Where would I start in finding the real and complex solutions to the equation: x³ + 1 = 0?

Factor: .(x + 1)(x² - x + 1) .= .0

Set each factor equal to 0 and solve.

. . x + 1 .= .0 . → . x = -1

. . x² - x + 1 .= .0 . . . This requires the Quadratic Formula.

. . . . . . . . . . .__________ . . . . . . . . _

. . . . . . .1 ± √1² - 4(1)(1) . . . . 1 ± i√3

. . x .= . --------------------- . = . ----------

. . . . . . . . . . 2(1) . . . . . . . . . . . .2

- Mar 31st 2007, 11:31 PMgretchen
Could you then say the solution is:

-1, cis(pi/3), cis(5pi/3)? - Apr 1st 2007, 05:35 AMSoroban
Hello, Gretchen!

Quote:

Could you then say the solution is: .-1, cis(π/3), cis(5π/3)?

- Apr 1st 2007, 07:56 AMThePerfectHacker
I want to spend some time explaining why this approach does not work.

Say we have,

x^2-1 = 0

We write

x^2 = 1

And take square roots,

x = 1.

But that is not true.

Because the statement sqrt(x^2)=x is false.

It should have been |x|=1

And then solve for "x".

Say we have,

x^4-1=0

Same thing,

x^4=1

|x|=1

But there is a problem.

|x| represens all complex numbers on the unit circle (have any idea what I am talking about?)

Thus, we want to find all complex numbers 1 unit away from the origin. The problem is there is an infinitude of such solutions. And we cannot check all of them. This is why this approach does not work. - Apr 3rd 2007, 02:57 PMSoroban
Hello, ThePerfectHacker!

Quote:

I want to spend some time explaining why this approach does not work.

Say we have: .x²-1 .= .0

We write: .x² = 1

And take square roots: .x = 1.

But that is not true.

Because the statement: sqrt(x²) = x is false.

It should have been: |x| = 1

And then solve for x.

Say, we have: .x^4 - 1 .= .0

Same thing: .x^4 = 1 . → . |x| = 1

But there is a problem.

|x| represents all complex numbers on the unit circle. .I disagree.

Thus, we want to find all complex numbers 1 unit away from the origin.

The problem is there is an infinitude of such solutions.

And we cannot check all of them.

This is why this approach does not work.

You are mixing absolute value of a real number

. . with the magnitude of a complex number.

If we solve: .x² = 9, we have: .|x| = 3

. . Then: .x = ±3 . . . and that's it.

But if we solve, say: .z^4 = 1

. . there are__four__fourth roots: .z .= .±1, ±i

These happen to be**on**a unit circle, but they are only four points.

. . They do not represent the entire circle.

You can write: .|z| = 1 . . . and this is true.

. . Each of the four roots is exactly one unit from the origin.

We take the*n*th root of 1, we get*n*roots . . . some real, some complex.

. . And all*n*of them lie on the unit circle.

They are the vertices of a regular n-gon.

- Apr 3rd 2007, 03:39 PMPlato
It is certainly true that there are unaccountably many points on the unit circle. However, viewing those points as complex numbers there are exactly four of them the fourth power of which is -1.

That is impossible to do, because “absolute value of a real number and the magnitude of a complex number” is one and the same concept. It absolute value is simply distance. |w| is the distance from zero for any complex number,w, real or not. - Apr 4th 2007, 05:54 PMThePerfectHacker
In my defense, I will quote Plato. Who said that "absolute value" is the same concept for both complex and real numbers.

Furthermore, I was showing necessary but not sufficient condtions. In order to solve an equation we must find the necessary conditions and show that they are sufficient.