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Home  >>  TN XII Math  >>  Complex Numbers
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Solve : $x^4+4=0$

This is the first part of the multi-part question Q4

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1 Answer

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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:
$x^4+4=0$
$x^4=-4$
$\Rightarrow x^4=4(\cos\pi+i\sin\pi)$
Therefore $x=4^{\large\frac{1}{4}}(\cos\pi+i\sin\pi)^{\large\frac{1}{4}}$
$\qquad\qquad=4^{\large\frac{1}{4}}(\cos(2k\pi+\pi)+i\sin(2k\pi+\pi))^{\large\frac{1}{4}}\;\;\;k\in z$
$\qquad\qquad=4^{\large\frac{1}{4}}(\cos(2k+1)\pi+i\sin(2k+1)\pi)^{\large\frac{1}{4}}\;\;\;k\in z$
$\qquad\qquad=\sqrt 2(\cos(2k+1)\large\frac{\pi}{4}$$+i\sin(2k+1)\large\frac{\pi}{4})$$\;\;\;k=0,1,2,3$
Step 2:
The roots are $\sqrt 2 cis\large\frac{\pi}{4},$$\sqrt 2 cis\large\frac{3\pi}{4},$$\sqrt 2 cis\large\frac{3\pi}{4},$$\sqrt 2 cis\large\frac{5\pi}{4},$$\sqrt 2 cis\large\frac{7\pi}{4}$
$\Rightarrow \sqrt 2 cis\large\frac{\pi}{4},$$\sqrt 2 cis\large\frac{3\pi}{4},$$\sqrt 2 cis\large\frac{3\pi}{4},$$\sqrt 2 cis(\large\frac{5\pi}{4}$$-2\pi),\sqrt 2 cis(\large\frac{7\pi}{4}$$-2\pi)$
$\Rightarrow \sqrt 2 cis\large\frac{\pi}{4},$$\sqrt 2 cis\large\frac{3\pi}{4},$$\sqrt 2 cis\large\frac{3\pi}{4},$$\sqrt 2 cis(\large\frac{-3\pi}{4}),$$\sqrt 2 cis(\large\frac{-\pi}{4})$
$\Rightarrow \sqrt 2 cis\big(\large\frac{\pm \pi}{4}\big),$$\sqrt 2cis\big(\large\frac{\pm 3\pi}{4}\big)$
answered Jun 13, 2013 by sreemathi.v
 

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