Q. The values of $(16)^{1/4}$ are
A) $\pm 2,\ \pm 2i$
B) $\pm 4,\ \pm 4i$
C) $\pm 1,\ \pm i$
D) None of these
Explanation
(a)
We have,
$(16)^{1/4} = \left( 2^{4} \right)^{1/4} = 2(1)^{1/4}$
$= 2\left( \cos{0 + i\sin 0} \right)^{1/4}$
$= 2(\cos{\frac{1}{4}(2k\pi + 0) + i\sin{\frac{1}{4}(2k\pi + 0)}}$
$k = 0,\ 1,\ 2,\ 3$
$= 2 \times 1,\ 2 \times i,\ 2 \times - 1,\ 2 \times - i$
$= \pm 2, \pm 2i$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.