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Question : 11 of 160
Marks:
+1,
-0
Solution:
We have,
‌x14+x9−x5−1=0‌⇒‌‌(x9−1)(x5+1)=0‌x9=1‌x=(1)1∕9‌x=(cos‌0∘.‌+isin‌0∘)1∕9‌=(cos‌2‌K‌π‌+isin‌2Kπ)1∕9‌=cos‌+isin‌‌using Demoivre's theorm here,
K=0,1,...,8‌x5=−1‌x=(cos‌π+isin‌π)1∕5‌=[cos(2Kπ+π)‌+isin‌(2Kπ+π)]1∕5‌=[cos(2K+1)‌π‌+isin‌(2K+1)π]1∕5‌=cos‌‌+isin‌‌‌‌ where ‌K=0,1,....5So, one of root of the given equation is
So, one of root
‌+isin‌‌ for
K=1=‌+i‌
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