Imagining Imaginary Numbers

Let $x$ be the answer to the expression.

Factoring the terms, we get

$\large x = \frac{(i^{32}+1)(i^{16}+1)(i^{8}+1)(i^{4}+1)(i^2+1)(i+1)}{(i^{2}+1)(i+1)}$

Now, we can cancel the term $i + 1$ , since its nonzero, but not $i^2 +1$ because it has a value of zero.

We see now that $x = \frac {0}{0}$ .

This implies that $x \times 0 = 0$ . Since any value of $x$ would make the statement correct, this makes it indeterminate .

PS: $x$ will only be undefined if $x = \frac {a}{0}$ for nonzero value of $a$ .

Also, we cannot use L'Hopital's rule here because $i$ is a number, and not a variable. After all, we're not looking for limits.

$\huge{=\frac { 1+i+{ i }^{ 2 }+{ i }^{ 3 }+{ i }^{ 4 }\dots { i }^{ 63 } }{ 1+i+{ i }^{ 2 }+{ i }^{ 3 } } \\ =\frac { 1+i+{ i }^{ 2 }+{ i }^{ 3 }+{ i }^{ 4 }\dots { i }^{ 63 } }{ 1+i+1-i } \\ =\frac { 1+i+{ i }^{ 2 }+{ i }^{ 3 }+{ i }^{ 4 }\dots { i }^{ 63 } }{ 0 } }$

Since the denominator is 0, we can immediately see that the fraction's value will be indeterminate. Hence there is no need to evaluate expression in the numerator.