MCQ
If $z = \frac{{\sqrt 3 + i}}{{ - 2}}$, then ${z^{69}}$ is equal to
- A$1$
- B$-1$
- ✓$i$
- D$-i$
✓
Answer
Correct option: C.
$i$
c
(c)$z = \frac{{\sqrt 3 + i}}{{ - 2}}$ $ \Rightarrow $ $iz = - \frac{{ - 1 + \sqrt {3i} }}{2} = - \omega $
$ \Rightarrow $ $z = \frac{{ - \omega }}{i} = i\omega $
(c)$z = \frac{{\sqrt 3 + i}}{{ - 2}}$ $ \Rightarrow $ $iz = - \frac{{ - 1 + \sqrt {3i} }}{2} = - \omega $
$ \Rightarrow $ $z = \frac{{ - \omega }}{i} = i\omega $
$ \Rightarrow $${z^{69}} = {i^{69}}.{\omega ^{69}} = i$
$(\because {\omega ^{3n}} = {i^{4n}} = 1)$
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