Complex Numbers And Quadratic Equations question 651
Question: The solution of $ 2\sqrt{2}x^{4}=(\sqrt{3}-1)+i(\sqrt{3}+1) $ is
Options:
A) $ \pm ( \cos \frac{5\pi }{48}+i\sin \frac{5\pi }{48} ) $
B) $ \pm ( \cos \frac{7\pi }{48}+i\sin \frac{7\pi }{48} ) $
C) $ \pm ( \cos \frac{19\pi }{48}-i\sin \frac{19\pi }{48} ) $
D) None of these.
Show Answer
Answer:
Correct Answer: A
Solution:
$ x^{4}=\frac{\sqrt{3}-1}{2\sqrt{2}}+i\frac{\sqrt{3}+1}{2\sqrt{2}}=\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12} $ So, $ x=\cos { \frac{k\pi }{2}+\frac{5\pi }{48} }+i\sin { \frac{k\pi }{2}+\frac{5\pi }{48} } $ $ k=0,1,2,3 $
$ \therefore $ Roots are $ \cos \frac{5\pi }{48}+isin\frac{5\pi }{48} fork=0 $ $ \cos \frac{29\pi }{48}+isin\frac{29\pi }{48} fork=1 $ $ \cos \frac{53\pi }{48}+isin\frac{53\pi }{48} =-( \cos \frac{5\pi }{48}+i\sin \frac{5\pi }{48} )fork=2 $ $ \cos \frac{77\pi }{48}+isin\frac{77\pi }{48} =-( \cos \frac{29\pi }{48}+i\sin \frac{29\pi }{48} ) $ for $ k=3 $
BETA
