We are given the expression:
\[
(-64i)^{\frac{5}{6}}
\]
Step 1: Convert to Polar Form
The given complex number is:
\[
-64i = 64 \times (-i)
\]
Since \( -i = e^{-i\pi/2} \), we rewrite:
\[
-64i = 64 e^{-i\pi/2}
\]
Expressing in polar form:
\[
r = 64, \quad \theta = -\frac{\pi}{2}
\]
Thus,
\[
-64i = 64 e^{-i\pi/2}
\]
Step 2: Apply Power Rule
Using De Moivre’s Theorem:
\[
(-64i)^{\frac{5}{6}} = 64^{\frac{5}{6}} e^{-i \frac{5\pi}{12}}
\]
Computing \( 64^{5/6} \):
\[
64^{\frac{5}{6}} = (2^6)^{\frac{5}{6}} = 2^{6 \times \frac{5}{6}} = 2^5 = 32
\]
Thus,
\[
(-64i)^{\frac{5}{6}} = 32 e^{-i \frac{5\pi}{12}}
\]
Step 3: Convert to Rectangular Form
Using Euler’s formula:
\[
e^{-i \frac{5\pi}{12}} = \cos\left(-\frac{5\pi}{12}\right) + i\sin\left(-\frac{5\pi}{12}\right)
\]
Since:
\[
\cos(-\theta) = \cos \theta, \quad \sin(-\theta) = -\sin \theta
\]
\[
e^{-i \frac{5\pi}{12}} = \cos \frac{5\pi}{12} - i\sin \frac{5\pi}{12}
\]
Approximating values:
\[
\cos \frac{5\pi}{12} = \frac{\sqrt{6} + \sqrt{2}}{4}, \quad \sin \frac{5\pi}{12} = \frac{\sqrt{6} - \sqrt{2}}{4}
\]
\[
e^{-i \frac{5\pi}{12}} = \frac{\sqrt{6} + \sqrt{2}}{4} - i\frac{\sqrt{6} - \sqrt{2}}{4}
\]
Multiplying by 32:
\[
32 \times \left(\frac{\sqrt{6} + \sqrt{2}}{4} - i\frac{\sqrt{6} - \sqrt{2}}{4} \right)
\]
\[
= 8( \sqrt{6} + \sqrt{2} ) - 8i( \sqrt{6} - \sqrt{2} )
\]
Factoring:
\[
= 16\sqrt{2} (1+i)
\]
Final Answer: \(\boxed{16\sqrt{2}(1+i)}\)
\bigskip