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Home  >>  TN XII Math  >>  Complex Numbers
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Find all the values of $\bigg(\large\frac{1}{2}$$-i\large\frac{\sqrt 3}{2}\bigg)^{\Large\frac{3}{4}}$ and hence prove that the product of the values is 1.

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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:
Let $x=\big(\large\frac{1}{2}-i\large\frac{\sqrt 3}{2}\big)^{\Large\frac{3}{4}}$
Now $\big|\large\frac{1}{2}$$-i\large\frac{\sqrt 3}{2}\big|=\frac{1}{4}+\frac{3}{4}$
$\qquad\qquad\quad\quad\;\;=1$
Step 2:
We have $\large\frac{1}{2}-$$i\frac{\sqrt 3}{2}$$=\cos\big(\large\frac{-\pi}{3}\big)$$+i\sin\big(\large\frac{-\pi}{3}\big)$
Therefore $x=[\cos\big(\large\frac{-\pi}{3}\big)$$+i\sin\big(\large\frac{-\pi}{3}\big)]^{\Large\frac{3}{4}}$
$\qquad\qquad\;\;=[\cos(-\pi)+i\sin(-\pi)]^{\large\frac{1}{4}}$
$\qquad\qquad\;\;=[\cos(2k\pi-\pi)+i\sin(2k\pi-\pi)]^{\large\frac{1}{4}}\;\;k\in z$
$\qquad\qquad\;\;=[\cos(2k-1)\pi+i\sin(2k-1)\pi]^{\large\frac{1}{4}}\;\;k\in z$
$\qquad\qquad\;\;=[\cos(2k-1)\large\frac{\pi}{4}$$+i\sin(2k-1)\large\frac{\pi}{4}$$]\;\;\;k=0,1,2,3$
Step 3:
The four values are $cis\big(\large\frac{-\pi}{4}\big),$$cis \;0,cis\large\frac{3\pi}{4},$$cis\big(\large\frac{5\pi}{4}\big)$
$\Rightarrow cis\big(\large\frac{-\pi}{4}\big),$$cis \large\frac{\pi}{4},$$\;cis\large\frac{3\pi}{4},$$cis\big(\large\frac{-3\pi}{4}\big)$

 

answered Jun 13, 2013 by sreemathi.v
edited Sep 12, 2013 by balaji.thirumalai
cis 0 should be changed to cis(π/4)
Thanks for pointing it out. It's fixed now.
 
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