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Solve : $x^4-x^3+x^2-x+1=0$

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

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  • 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:
Multiplying by $4(x+1)$ we have the equation
The roots of equ(1) are the same as those of equ(2) except for $x=-1$
Step 2:
Solving eq(2) we have $x=(-1)^{\large\frac{1}{5}}$
$\;\;=(\cos2k\pi+\pi)+i\sin(2k\pi+\pi))^{\large\frac{1}{5}}\;\;\;k\in z$
$\;\;=[\cos(2k+1)\pi+i\sin(2k+1)\pi)]^{\large\frac{1}{5}}\;\;\;k\in z$
Step 3:
Th value $k=2$ gives $x=\cos\pi+i\sin\pi=-1$
Which is not a root of eq(1).
The roots of eq(1) are $cis\large\frac{\pi}{5},$$cis\large\frac{3\pi}{5},$$cis\large\frac{7\pi}{5},$$cis\large\frac{9\pi}{5}$
$\Rightarrow cis\large\frac{\pi}{5},$$cis\large\frac{3\pi}{5},$$cis(\large\frac{7\pi}{5}-$$2\pi),$$cis(\large\frac{9\pi}{5}$$-2\pi)$
$\Rightarrow cis\big(\pm\large\frac{ \pi}{5}\big),$$cis\big(\pm\large\frac{ 3\pi}{5}\big)$


answered Jun 13, 2013 by sreemathi.v
edited Sep 12, 2013 by balaji.thirumalai
cis(±±π/5) should be changed to cis(±π/5)
Thanks for pointing it out, its fixed now.

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