Question

One of the values of \( (-64i)^{5/6} \) is:}

• A. \( 32i \)
• B. \( 16\sqrt{2}(1+i) \)
• C. \( 32(1+i) \)
• D. \( 16\sqrt{2} \) \bigskip

Hint:

To raise a complex number to a fractional power, express it in polar form first, then apply De Moivre's Theorem to find the result.

Answer 1

The Correct Option is B

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