How to find a real function from a complex function. I have the complex function $z\left(n\right) = i^{n} = \cos\left(\theta\left(n\right)\right) + i \sin\left(\theta\left(n\right)\right), \theta\left(n\right) = \frac{n \pi}{2},$ and I know that, on an Argand diagram, it would trace the function $x^{2} + y^{2} = 1,$ but I would like to know why this is the case.
Also, is there any way to convert a complex function into a real function, in that the real function would trace the complex function on the real plane, for any complex function?
I am sure that it must be possible to, but I do not know how.
 A: 
...and I know that, on an Argand diagram, it would trace the function $x^2+y^2=1,$ but I would like to know why this is the case.

First I will assume $n$ is an integer.  Then powers of $i$ are periodic, repeating every 4 integers.  The sequence  $\{i^n\}$ begins (starting at $n=0$) $\{1,i,-1,-i,\ldots\}.$  So it is therefore natural to represent these numbers on the real and imaginary axes in the complex plane where the modulus of each number is 1.  Hence the Argand diagram.
The graph of $x^2+y^2=1$ in the real plane also has the characteristic that all points are 1 unit from the origin, but when you are talking about this relation, you are in the real plane, and so you are only seeing the modulus of your complex-valued function.

...is there any way to convert a complex function into a real function, in that the real function would trace the complex function on the real plane, for any complex function?

The answer to this is 'no,' but you can graph the modulus of a complex function, which is real-valued.
So for your example, $z(n)=i^n$ in the complex plane includes points along the path of $x^2+y^2=1$ in the real plane because
$$|z(n)|=|x^2+y^2|\text{ for all }z=x+iy\in\mathbb{C}.$$
