MCQ
If $i{z^4} + 1 = 0$, then $z$ can take the value
- A$\frac{{1 + i}}{{\sqrt 2 }}$
- ✓$\cos \frac{\pi }{8} + i\,\sin \frac{\pi }{8}$
- C$\frac{1}{{4i}}$
- D$i$
✓
Answer
Correct option: B.
$\cos \frac{\pi }{8} + i\,\sin \frac{\pi }{8}$
b
(b) $i{z^4} = - 1$
${z^4} = \frac{{ - 1}}{i} \Rightarrow {z^4} = i \Rightarrow z = {(i)^{1/4}}$
$z = {(0 + i)^{1/4}}$
$z = {\left( {\cos \frac{\pi }{2} + i\sin \frac{\pi }{2}} \right)^{1/4}}$
$z = \cos \frac{\pi }{8} + i\sin \frac{\pi }{8}$ (using De Moivre’s theorem)
(b) $i{z^4} = - 1$
${z^4} = \frac{{ - 1}}{i} \Rightarrow {z^4} = i \Rightarrow z = {(i)^{1/4}}$
$z = {(0 + i)^{1/4}}$
$z = {\left( {\cos \frac{\pi }{2} + i\sin \frac{\pi }{2}} \right)^{1/4}}$
$z = \cos \frac{\pi }{8} + i\sin \frac{\pi }{8}$ (using De Moivre’s theorem)
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