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Home  >>  TN XII Math  >>  Complex Numbers
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Find all the value of the following : $(8i)^{\large\frac{1}{3}}$

This is the second part of the multi-part question Q1

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Toolbox:
  • From De moivre's theorem we have
  • (i) $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta,n\in Q$
  • (ii) $(\cos\theta+i\sin\theta)^{-n}=\cos n\theta-i\sin n\theta$
  • (iii) $(\cos\theta-i\sin\theta)^n=\cos n\theta-i\sin n\theta$
  • (iv) $(\sin \theta+i\cos \theta)^n=[\cos(\large\frac{\pi}{2}$$-\theta)+i\sin(\large\frac{\pi}{2}$$-\theta)]^n=\cos n(\large\frac{\pi}{2}$$-\theta)+i\sin n(\large\frac{\pi}{2}$$-\theta)$
  • $e^{i\theta}=\cos\theta+i\sin\theta$
  • $e^{-i\theta}=\cos\theta-i\sin\theta$,also written as $\cos\theta$ and $\cos(-\theta)$
Step 1:
$(8i)^{\large\frac{1}{3}}=2^{\large\frac{1}{3}}i^{\large\frac{1}{3}}$
$\Rightarrow 2^{\large\frac{1}{3}}[\cos\large\frac{\pi}{2}$$+i\sin\large\frac{\pi}{2}]^{\large\frac{1}{3}}$
$\Rightarrow 2^{\large\frac{1}{3}}[\cos(2k\pi+\large\frac{\pi}{2})$$+i\sin(2k\pi+\large\frac{\pi}{2})]^{\large\frac{1}{3}}$$\;\;k\in z$
$\Rightarrow 2^{\large\frac{1}{3}}[\cos(4k+1)\large\frac{\pi}{2}+$$i\sin(4k+1)\large\frac{\pi}{2}]^{\large\frac{1}{3}}$$\;\;\;k\in z$
$\Rightarrow 2^{\large\frac{1}{3}}[\cos(4k+1)\large\frac{\pi}{6}+$$i\sin(4k+1)\large\frac{\pi}{6}]$$\;\;\;k=0,1,2$
Step 2:
The roots are $2^{\large\frac{1}{3}}(\cos\large\frac{\pi}{6}$$+i\sin\large\frac{\pi}{6}),$$2^{\large\frac{1}{3}}$$(\cos\large\frac{5\pi}{6}+$$i\sin\large\frac{5\pi}{6}),$$2^{\large\frac{1}{3}}$$(\cos\large\frac{3\pi}{2}$$+i\sin\large\frac{3\pi}{2})$
$\Rightarrow 2^{\large\frac{1}{3}}(\large\frac{\sqrt 3}{2}+\frac{i}{2})$$,2^{\large\frac{1}{3}}(\large\frac{-\sqrt 3}{2}+\frac{i}{2})$$,-2^{\large\frac{1}{3}}i$
answered Jun 12, 2013 by sreemathi.v
 
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