We have to solve the following task of determing the sum and products of all sixth unit roots.
I have started as follows:
$z^6 = 1 \Longleftrightarrow (z^3-1)(z^3+1) = 0 \Longleftrightarrow (z-1)(z^2+z+1)*(z+1)(z^2-z+1) = 0$
And hence the roots would be $1$,$-1$, $(-1)^{2/3}$ and $-(-1)^{2/3}$. Is this correct so far?
How comes that in forming $z^6 = 1$ to $(z^3-1)(z^3+1) = 0$ I can treat z just like a real number also in "interactions" with real numbers (such as "1" here)?
it would be better if you can try considering $(cos \theta +i \sin \theta)^6=1$
i.e., $(cos \theta +i \sin \theta)^6=1 \rightarrow \cos 6\theta + i\sin 6\theta =1$
i.e., $\cos 6\theta =1\Rightarrow 6\theta =?? \Rightarrow \theta =??$
If you know what $\theta$ is (make sure you get $6$ different values), then you can just write
$cos \theta +i \sin \theta$ for those values of $\theta $..
Good luck!
The simple answer to your question is that all real numbers are also complex numbers - they're just complex numbers with imaginary part $0$.
In general, however, you can do a lot of the sort of algebra you're used to with real numbers in much more general settings. To give you an example - in mathematics, we have something called a field, which is a set $K$ (think of this as just a collection of objects, such as the collection of all real numbers) together with two operations $+$ and $\times$ which behave sort of the way that $+$ and $\times$ do on the real numbers. That is to say, if $x,y,z$ are elements in $K$, they satisfy the field axioms:
\begin{align} (x+y)+z&=x+(y+z)\\ x+y&=y+x\\ (xy)z&=x(yz)\\ xy&=yx\\ x(y+z)&=xy+xz\\ \end{align}
There are a few more field axioms: two specify the existence of identity elements, which are numbers which we'll call $0$ and $1$ which behave in the same way as the real numbers $0$ and $1$: if $x$ is an element of $K$ then:
\begin{align} x+0&=x\\ x\times1&=x\\ \end{align}
The other two axioms specify inverse elements - these allow us to do subtraction and division. Specifically, if $x$ is in $K$, then there is a number in $K$ which we'll call $-x$ and (if $x\ne0$) a number in $K$ which we'll call $x^{-1}$, such that:
\begin{align} x+-x&=0\\ x\times x^{-1}&=1 \end{align}
The condition that $x^{-1}$ only exists if $x$ is not equal to $0$ is in there for a reason, which the following exercise demonstrates.
Exercise: show that if $0^{-1}$ exists, and $x,y$ are members of $K$, then $x=y$.
In other words, the only field in which $0^{-1}$ exists is the rather strange field with only one element. Most authors specify that you have to have $1\ne0$, so this is not a field at all.
The field axioms are automatically enough to prove several properties analogous to those in the real numbers. For example:
Exercise: show, from the axioms above, that if $x$ is a member of $K$ then: $$ x\times0=0 $$
Now it should be fairly clear that the real numbers are an example of a field. It should also be fairly easy to see that the complex numbers are also an example of a field. There are many others.
The nice thing about the field axioms is that all the algebraic operations of addition, multiplication, subtraction and division that you're familiar with from real algebra carry over to them.
Exercise: Let $K$ be a field, and denote by $2$ the number $1+1$. If $x$ is a member of the field, we can write $x^2$ to mean $x\times x$. Show that, if $x,y$ are in $K$, then $(x+y)^2=x^2+2xy+y^2$.
That's a reflection that, when we show that $(x+y)^2=x^2+2xy+y^2$ in the real numbers, all we're using about the real numbers are that they're a field under the operations $+$ and $\times$. It also provides an example to show that we can do many things in fields that are very similar to what we do in the real numbers. For example, we can define an element of any field called $2$ by $2=1+1$, even if the field is not the real numbers. We can define $3=2+1,4=3+1$ etc. The only thing to be careful about is that some of these numbers might be equal to one another. For example, in most fields, you can define a number $\dfrac12$ to be $2^{-1}=(1+1)^{-1}$ (which might be useful for completing the square, say), but in some fields (such as the field of integers$\mod 2$, where the only two elements are $0$ and $1$ and $1+1=0$), $2=0$, so it doesn't have a multiplicative inverse.
Anyway, in your example, all you're using are the field properties of the complex numbers, which are the same as those of the real numbers. It's a bit of work to check that the axioms all hold for the complex numbers, I suppose, but they do, and once you've done that check you'll find that a lot of the real number algebra you're used to in fact carries over to all fields.
