# Thread: De Moivre's Theorem of Complex numbers Question Help

1. ## De Moivre's Theorem of Complex numbers Question Help

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??

2. ## Re: De Moivre's Theorem of Complex numbers Question Help

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

3. ## Re: De Moivre's Theorem of Complex numbers Question Help

Originally Posted by BobP
The LHS will factorise. Then you can equate each factor to zero.
And it factorises to...

\displaystyle \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 \displaystyle \begin{align*} z^5 - 1 = 0 \end{align*} and \displaystyle \displaystyle \begin{align*} z^4 + 1 = 0 \end{align*} and solve using DeMoivre's Theorem

4. ## Re: De Moivre's Theorem of Complex numbers Question Help

Originally Posted by Prove It
And it factorises to...

\displaystyle \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 \displaystyle \begin{align*} z^5 - 1 = 0 \end{align*} and \displaystyle \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