TS EAMCET · Maths · Complex Number
If \(Z=r(\cos \theta+i \sin \theta),(\theta \neq-\pi / 2)\) is a solution of \(x^3=i\), then \(r^9(\cos \theta+i \sin \theta)^9=\)
- A \(\frac{\sqrt{3}}{2}+\frac{1}{2} i\)
- B 1
- C -i
- D \(\frac{-\sqrt{3}}{2}+\frac{1}{2} i\)
Answer & Solution
Correct Answer
(C) -i
Step-by-step Solution
Detailed explanation
\(Z^3 = i\) \(r^3(\cos 3\theta+i \sin 3\theta) = \cos(\pi/2)+i \sin(\pi/2)\) \(r^3=1 \Rightarrow r=1\) \(3\theta = \pi/2 + 2k\pi\) \(r^9(\cos \theta+i \sin \theta)^9 = 1^9(\cos 9\theta+i \sin 9\theta)\) \(\cos(3(3\theta))+i \sin(3(3\theta))\)…
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