Find real/complex zeros

Question

what are the real/complex zeros for:

$t^9 - 1$

I also need to use the exponential form of complex numbers

Answer 1

\begin{align} t^9-1&=(t^3-1)(t^6+t^3+1)\\ &=(t-1)(t^2+t+1)(t^6+t^3+1) \end{align}

The complex zeroes can be solved by means of quadratics.

Answer 2

$\exp\left(\dfrac{2\pi i k}9\right)$ for $k=1$ to $9$.

That's real only for $k=9$.

Answer 3

The truly WONDERFUL thing about exponential form of complex numbers is that if

$z = re^{i\theta}$ and $w = se^{i\phi}$ then $z\cdot w = (rs)e^{i(\theta + \phi \pm\text{some multiples of }2\pi\text{ to take care of cases where angles swing around a full circle})}$

This means if $z^m = r^me^{i(m\theta\pm\text{some multiples of }2\pi\text{ to take care of cases where angles swing around a full circle})}$

And it means we can solve: if $z^n = re^{i\theta}$ then the $n$ roots are:

$\sqrt[n]{r} e^{i(\frac \theta n + \text{some multiples }\frac {2\pi}k\text{ to take care of cases where angles swung around a full circle})}$

or in other words, the possible values for $z$ are

$\sqrt[n]{r} e^{i(\frac \theta n +\frac {2\pi}k)}$ for $k=0....., n-1$.

So since $1 = 1*e^{i*0}$ then if $z^9 = 1$ then the nine possible values are:

$\sqrt[9]{1}*e^{i(\frac 09 + \frac {2\pi}k} = e^{i(0\frac {i2\pi}k}=$

Or the $9$ roots are $e^{i0} =1; e^{i\frac 29\pi}; e^{i\frac 39\pi};.... e^{i\frac 89\pi}$.

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Now if you are like nearly every person first seeing this you are probably confuse be the $\frac {2\pi}k$ and the "$\pm\text{some multiples of }2\pi\text{ to take care of cases where angles swing around a full circle}$" business.

The this is the if we have a number such as $z = e^{i\theta}$ where $\pi < \theta < 2\pi$ then $z^2 = e^{i\theta}\cdot e^{i\theta}= e^{i2\theta}$. But $2\pi < 2\theta$ so the angle has swung "full circle". And $e^{i2\theta} = e^{i(2\theta - \2pi)}$. And if we let $\eta = 2\theta - 2\pi$ then $z^2 = e^{i\eta}$.

SO if we were given $w= e^{i\eta}$ at the beginning, and asked: find $z$ so that $z^2 = e^{i\eta}$ we would figure $z = e^{i\frac {\eta}2}$. That's fine and that is one of the roots (Notice that $0< \eta<\pi$) but we need to take into account what happens when the angle in $z$ is large enough that doubling it will "swing it past a full circle".

In other words: $w = e^{i\eta} = e^{i(\eta + 2\pi)}$ so when we have $z^2 = w$ we have that $z$ can be $e^{i\frac {\eta} 2}$ or $z$ can be $e^{i\frac {\eta+2\pi}2}=e^{i(\frac \eta 2 + \pi)}$. Notice $\frac \eta 2 + \pi = \theta$, our original big angle. And Notice that $\frac \eta 2 =\theta-\pi$.