Author Topic: z^4+z^3+z^2+z+1=0 (Read 1846 times) Tweet Share

/0 · « **on:** February 16, 2009, 08:55:25 am »

Can someone please find the solutions to $z^4+z^3+z^2+z+1=0$ . I've been looking online but to no avail. Surely someone out there must have found the soluions.

(in b4 use quartic formula)

sb3700 · « **Reply #1 on:** February 16, 2009, 11:58:42 am »

One way to do it:

Note that the coefficients of z^4/z^0, and z^3/z are the same. So we can use this trick.

Clearly $z=0$ is not a solution.
So we can divide by $z^2$ both sides giving:
$z^2 + z + 1 + 1/z + 1/z^2 = 0$

Note that $(z + 1/z)^2 = z^2 + 1/z^2 + 2$
So $z^2 + 1/z^2 = (z + 1/z)^2 - 2$

Sub this in above:
$0 = (z + 1/z)^2 + (z + 1/z) - 1$
Put $X = z + 1/z$
So $5/4 = (X + 1/2)^2$ (complete the square)
And $X = z + 1/z = \frac{-1 \pm \sqrt(5)}{2}$

So $z^2 + 1 - \frac{-1 \pm \sqrt(5)}{2} z = 0$

Solve those 2 cases for 4 complex solutions. Normally, there are better numbers involved when you are doing it by hand.

For example, if it was $z^4 + 3z^3 - 2z^2 + 3z + 1 = 0$
Then it simplifies to $(z + 1/z)^2 + 3(z + 1/z) - 4 = 0$
So $z + 1/z = -4$ or $z + 1/z = 1$
Then $z^2 + 4z + 1 = 0$ or $z^2 - z + 1 = 0$
$(z + 2)^2 = 3$ or $(z - 1/2)^2 = -3/4$
So $z = -2 \pm \sqrt(3)$ or $z = \frac{1 \pm \sqrt(3) i}{2}$

Hopefully there are no errors in the working.

/0 · « **Reply #2 on:** February 16, 2009, 12:56:30 pm »

Wow, nice solution sb3700

Thankyou

In the meant time I think I worked out another method

$z^4+z^3+z^2+z+1=\frac{1(z^5-1)}{z-1}=0$ . Since it is a geometric series...

So $z^5-1=0$ for $z\neq 1$ and we can see $z=1$ is not a solution.

Therefore we have reduced $z^4+z^3+z^2+z+1=0$ to solving $z^5=1$ , $z\neq 1$ , which I think is pretty neat. :laugh:

I was trying to solve equation of the form $z^n+z^{n-1}+...+z^0=0$ to see their patterns.

So far it seems like $\sum_{i=0}^n z^i=0$ has solution which form an $n+1$ -gon, except that there is a point missing in each case, $z=1$ .

sb3700 · « **Reply #3 on:** February 16, 2009, 01:14:40 pm »

Yours is better I'd say, and allows you to solve it for your whole class of problems.

Quote from: /0 on February 16, 2009, 12:56:30 pm

So far it seems like $\sum_{i=0}^n z^i=0$ has solution which form an $n+1$ -gon, except that there is a point missing in each case, $z=1$ .

Yeah. $z^{n+1} - 1 = (z-1) * (z^n + \cdots + 1)$
So $\sum_{i=0}^n z^i = 0$ implies $\frac{z^{n+1} - 1}{z-1} = 0$ , $z \ne 1$

/0 · « **Reply #4 on:** February 16, 2009, 06:36:55 pm »

Hmmm yeah but your solution took skill whereas mine was a lucky guess.

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Author Topic: z^4+z^3+z^2+z+1=0 (Read 1846 times) Tweet Share

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z^4+z^3+z^2+z+1=0

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