Real Numbers to Complex?!

Since x can only be a non-zero real number we can solve this with simple but clever casework:

Case 1: (0 < x)

Case 2: (0 > x)

Case 3: (1 > x > 0)

In this three cases, if we substitute we get a non-zero real number, however for case 4, we get two subcases which vary on the parity of x’s in the equation:

Case 4: (0 > x > -1)

Subcase 1: (the number of x’s is even)

If the number of x’s is even, the number of x’s that are exponents will be uneven: We can define x as - $\frac{r}{p}$

The equation would be:

$\left(-\frac{r}{p}\right)^{\frac{1}{\frac{1}{\frac{1}{\left(-\frac{r}{p}\right)^{\left(-\frac{r}{p}\right)^{\left(-\frac{r}{p}\right)^{.^{.^{.}}}}}}}}}$ = $\left(-\frac{r}{p}\right)^{\frac{1}{\left(-\frac{r}{p}\right)^{.^{.^{.}}}}}$

Now, we’re elevating x to the x an indefinate amount of times, so in the case that we have an even amount of x’s (uneven amount for x’s that are part of the exponent) we can be sure that the exponent will be negative. Now let’s define $\left(-\frac{r}{p}\right)^{.^{.^{.}}}$ as $-\frac{h}{k}$ (Note: It’s kept keep it as a rational term, because every power of a fraction whose denominator is different to one is another fraction, and it’s negative because when we elevate a negative base to an uneven power it will remain negative, also, when we elevate an uneven number to an uneven power we get another uneven number).

Finally, we can say

$\left(-\frac{r}{p}\right)^{\frac{1}{\left(-\frac{r}{p}\right)^{.^{.^{.}}}}}$ = $\left(-\frac{r}{p}\right)^{\frac{1}{-\frac{h}{k}}}$ = $\left(-\frac{r}{p}\right)^{-\frac{k}{h}}$ This can be rewritten as:

$\sqrt[h]{\left(-\frac{r}{p}\right)^{-k}}$ = $\sqrt[h]{\left(-\frac{p}{r}\right)^{k}}$

Here we can see that a complex number is possible, especifically when k is an odd number. $\Box$