Hi everyone

I'm trying to solve this equation:

z^{4}+1 = 0

I've tried every method I can think of, such as De Moivre's formula, but I only get two complex roots.

What method do I use ?

I just want to know the method/procedure.

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- September 17th 2012, 10:41 AMTalaPolynomial with real coefficients but complex roots
Hi everyone

I'm trying to solve this equation:

z^{4}+1 = 0

I've tried every method I can think of, such as De Moivre's formula, but I only get two complex roots.

What method do I use ?

I just want to know the method/procedure. - September 17th 2012, 10:49 AMMarkFLRe: Polynomial with real coefficients but complex roots
You could rewrite the equation as:

- September 17th 2012, 10:52 AMTalaRe: Polynomial with real coefficients but complex roots
Should I still use De Moivre's formula ? Or ?

- September 17th 2012, 11:08 AMMarkFLRe: Polynomial with real coefficients but complex roots
You could use Euler/de Moivre (which are easier), but another method would be to set:

By equating coefficients, you may determine the 4 roots.

Using Euler and de Moivre:

where

Can you finish? - September 17th 2012, 11:19 AMTalaRe: Polynomial with real coefficients but complex roots
I have to be honest and say that both methods seem a bit unclear to me.

- September 17th 2012, 11:28 AMMarkFLRe: Polynomial with real coefficients but complex roots

Now, use the 4 values we found above for , i.e., . - September 17th 2012, 11:30 AMTalaRe: Polynomial with real coefficients but complex roots
Can you explain step 1 in Euler and De Moivre ?

- September 17th 2012, 11:44 AMMarkFLRe: Polynomial with real coefficients but complex roots
We have:

We find: hence, we may state:

So, we must have:

where

By Euler's formula, we know then:

Now, we want:

Since*k*is an integer, we have

Hence, we have:

Now, use the values we found for*k*to compute the 4 roots. - September 17th 2012, 11:51 AMTalaRe: Polynomial with real coefficients but complex roots
I understand now !

Thank you so much - September 17th 2012, 02:09 PMSorobanRe: Polynomial with real coefficients but complex roots
Hello, Tala!

Quote:

DeMoivre's Theorem is the approach I would use.

We have: .

Hence: . .

n . . . . . .