De Moivre's Theorem of Complex numbers Question Help

**donny05** · May 26th 2012, 06:54 AM

Hi I'm having trouble solving this complex number problem. If anyone can help me out i will really appreciate it. Thanks

here is the question
Use De Moivre's Theorem to solve this equation z⁹ + z⁵ - z⁴ -1 = 0?

I just want to know how can i start solving this??

**BobP** · May 26th 2012, 07:07 AM

The LHS will factorise. Then you can equate each factor to zero.

**Prove It** · May 26th 2012, 07:18 AM

Originally Posted by BobP

The LHS will factorise. Then you can equate each factor to zero.

And it factorises to...

$\displaystyle \begin{align*} z^9 + z^5 - z^4 - 1 &\equiv z^9 - z^4 + z^5 - 1 \\ &\equiv z^4\left(z^5 - 1\right) + 1\left(z^5 - 1\right) \\ &\equiv \left(z^5 - 1\right)\left(z^4 + 1\right) \end{align*}$

So that means you can set $\displaystyle \begin{align*} z^5 - 1 = 0 \end{align*}$ and $\displaystyle \begin{align*} z^4 + 1 = 0 \end{align*}$ and solve using DeMoivre's Theorem

**donny05** · May 26th 2012, 08:17 AM

Originally Posted by Prove It

And it factorises to...

$\displaystyle \begin{align*} z^9 + z^5 - z^4 - 1 &\equiv z^9 - z^4 + z^5 - 1 \\ &\equiv z^4\left(z^5 - 1\right) + 1\left(z^5 - 1\right) \\ &\equiv \left(z^5 - 1\right)\left(z^4 + 1\right) \end{align*}$

So that means you can set $\displaystyle \begin{align*} z^5 - 1 = 0 \end{align*}$ and $\displaystyle \begin{align*} z^4 + 1 = 0 \end{align*}$ and solve using DeMoivre's Theorem

Thanks alot!
i am able to solve it now, it was pretty simple. thanks again

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