Answer
$i$
Work Step by Step
Use exponent rule $a^{-b}=\dfrac{1}{a^b}$ to obtain:
$\begin{align*}
i^{-15}&=\dfrac{1}{i^{15}}\\
&=\dfrac{1}{i^{14}\cdot i}\\
&=\dfrac{1}{(i^{2})^7\cdot i}\\
\end{align*}$
Use $i^2=-1$.
$=\dfrac{1}{(-1)^7\cdot i}$
$=\dfrac{1}{-1\cdot i}$
$=\dfrac{1}{-i}$
Multiply both the numerator and the denominator by the conjugate of $-i$ which is $i$ to obtain:
$=\dfrac{1}{-i}\cdot \dfrac{i}{i}$
Simplify.
$=\dfrac{i}{-i^2}$
Use $i^2=-1$.
$=\dfrac{i}{-(-1)}$
$=\dfrac{i}{1}$
$=i$
Hence, the solution in the standard form is $i$.
