Perplexed Number

We have ${ i }^{ -9999 }=\frac { 1 }{ { i }^{ 9999 } } =\frac { 1 }{ { i }^{ 4\cdot 2499+3 } } =\frac { 1 }{ { i }^{ 4\cdot 2499 }\cdot { i }^{ 3 } } .$

Since ${ i }^{ 4 }=1$ , the equation can be simplified as

$\frac { 1 }{ { i }^{ 3 } } =\frac { 1 }{ -i } =-\left( \frac { 1 }{ i } \right) =-\left( -i \right) =i .\square$

$i^{-9999} = i^{-9999} * 1^{2500} = i^{-9999} * (i^4)^{2500} = i^{-9999 + 10000} = i$

$i^{-9999} = \frac{1}{i^{9999}} = \frac {i}{i^{10000}} = i$

i^4 = 1, so i ^10000 = 1. i^(-9999) = i^(-9999)*i^10000 = i^1 = i.

$\large i^{-9999}=i^{-9999+10000}=i.$ Adding or subtracting any multiple of four does not change the value. So, add or subtract to/from the exponent, such a number that the result is 0, 1, 2, or 3 and we get the answer. OR $For~~i^n ~find~n\equiv ~x~ (mod~4), i^n=i^x$ if you know mods.

Consider $-9999 \equiv -3 \ (mod \ 4)$ Which is no other than 1

Note that the powers of $i$ cycle ever $4$ powers. This means that the value of $i$ depends on the module $4$ of the power, where module $0$ is just $1$ . So, we have that $i^{-9999}=i^{-10000}\cdot i=1\cdot i=\boxed{i}$

• Let $\frac{1}{i}$ =x
• Then, 1=x $i$
• $i^4$ =x $i$
• x= $i^3$
• $\frac{1}{i}$ = $i^3$
• Therefore, $\frac{1}{(i^-9999)}$ = $i$