Answer
$i$
Work Step by Step
Use the exponent rule $a^{-b}=\frac{1}{a^b}$.
$i^{-23}\\
\\=\dfrac{1}{i^{23}}$
$=\dfrac{1}{i^{22}\cdot i}$
$=\dfrac{1}{i^{2(11)}\cdot i}$
$=\dfrac{1}{(i^{2})^{11}\cdot i}$
Use $i^2=-1$.
$=\dfrac{1}{(-1)^{11}\cdot i}$
$=\dfrac{1}{-1\cdot i}$
$=\dfrac{1}{-i}$
Multiply both the numerator and the denominator by the conjugate of $-i$ which is $i$.
$=\dfrac{1}{-i}\cdot \dfrac{i}{i}$
Simplify.
$=\dfrac{i}{-i^2}$
Use $i^2=-1$.
$=\dfrac{i}{-(-1)}$
$=\dfrac{i}{1}$
$=i$
Hence, the solution in the standard form is $i$.
