© examsiri.com
Question : 8 of 160
Marks:
+1,
-0
Solution:
Given,
(√3−i)n,n∈N‌‌ Modulus ‌|√3−1|‌=√(√3)2+(−1)2‌=√3+1=√4=2‌θ=arg(√3−i)‌=tan−1(‌)=‌Argument :
θ=arg(√3−i)=tan−1(‌)=‌So, in polar form,
‌‌√3−i=2⋅(cos‌θ+isin‌θ)‌‌=2[cos(‌)+sin‌(‌)]‌‌=2[cos‌−isin‌‌]‌⇒(√3−i)n={2⋅[cos(‌)−isin‌(‌)]}n‌‌=2n⋅[cos(‌)−isin‌(‌)].But given that
(√3−1)n=2nFrom Eqs. (i) and (ii), we get
‌2n[cos(‌)−isin‌(‌)]=2n⇒cos(‌)−isin‌(‌)=1⇒cos(‌)=1‌ and ‌sin‌(−‌)=1⇒−‌=2kπ⇒n=‌=−12kSince,
n∈N, for the smallest positive
n, let
k=−1⇒‌‌n=−12(−1)=12∴ The least positive value of
n is 12 .
© examsiri.com
Go to Question: