Question about Geometric Series and Roots of Unity

Hello. I am working on my own through the AQA FP2 Mathematics textbook (A2-Level) and I have come across a question I am stuck on. I am given the following equation, where $\displaystyle z$ is a complex number, and asked to solve the equation, in other words find the roots of the equation. My answers should be in the form $\displaystyle a + bi$.

$\displaystyle 1 - 2z + 4z^2 - 8z^3 = 0$

Initially I noticed that this is a geometric series and can therefore be expressed as a summation:

$\displaystyle \sum (-2z)^k$ from $\displaystyle k=0$ to $\displaystyle 3$

But using the related formula for a geometric series I get:

$\displaystyle \frac{a(1-r^n)}{1-r}$ = $\displaystyle \frac{-2(1-z^4)}{1-z}$

note: n=4 because the final term of a sum of a geometric series is always n-1 (3).

However, when I substitute a random number in for $\displaystyle z$, say 2, the two expressions do not equate. So somewhere, my geometric series expression is incorrect, but I cannot see where. I have not done geometric series in a while.

Continuing anyway, I would then go on to say that the roots of the initial equation must be three of the roots of $\displaystyle z^4 - 1 = 0$ or $\displaystyle z^4 = 1$. But when finding the four solutions to this equation (disregarding the solution $\displaystyle z=1$, as this would make the fraction indefinable), I do not get the right solutions.

To help, the actual solutions are $\displaystyle z = \frac{1}{2}, \pm \frac{1}{2}i$.

If anyone could help me see what I have done wrong with the geometric series, I would be very grateful.

Thanks in advance.

Re: Question about Geometric Series and Roots of Unity

why should we complicate the situation by going in for GP. Just group the terms and we get

( 1 - 2z ) + 4z^2 ( 1 - 2z ) = ( 1-2z)( 1 + 4z^2)=0

that gives z = 1/2 and z^2= -1/4 thus z = + 0r - i/2

Now for the GP

The formula is right

Sum = { 1 - ( -2z)^4}/ { 1 - ( -2z)} = ( 1 - 16z^4)/(1+2z) = { ( 1+4z^2)(1+2z)(1-2z) / ( 1+ 2z) } = ( 1+4z^2)(1-2z) That is the same what we got earlier.

Re: Question about Geometric Series and Roots of Unity

Thank you very much. Yeah, when I did the sum, I didn't realise 'r' was '(-2z)' and not just 'z' with -2 factored out. That makes sense. :)

Re: Question about Geometric Series and Roots of Unity

If you are going to sum the geometric series, get your a and r right. Your $\displaystyle \displaystyle \begin{align*} a = 1 \end{align*}$ and your $\displaystyle \displaystyle \begin{align*} r = -2z \end{align*}$, so the sum is actually $\displaystyle \displaystyle \begin{align*} \frac{a \left( 1 - r^n \right) }{1 - r} = \frac{1 \left[ 1 - \left( -2z \right) ^4 \right] }{1 - (-2z) } = \frac{1 - 16z^4}{1 + 2z} \end{align*}$.

So solving your equation:

$\displaystyle \displaystyle \begin{align*} 1 - 2z + 4z^2 - 8z^3 &= 0 \\ \frac{1 - 16z^4}{1 + 2z} &= 0 \\ 1 - 16z^4 &= 0 \\ \left( 1 - 4z^2 \right) \left( 1 + 4z^2 \right) &= 0 \\ \left( 1 - 2z \right) \left( 1 + 2z \right) \left( 1 - 2i\,z \right) \left( 1 + 2i\,z \right) &= 0 \\ z = \left\{ \frac{1}{2}, -\frac{1}{2}, -\frac{i}{2} , \frac{i}{2} \right\} \end{align*}$

But since $\displaystyle \displaystyle \begin{align*} z \neq -\frac{1}{2} \end{align*}$ as it would give a 0 denominator, we must disregard this answer.