© examsiri.com
Question : 8 of 160
Marks:
+1,
-0
Solution:
‌(‌)n=−1‌∵‌ Modulus of numerator ‌=√(√3)2+12‌=2θ=tan−1(‌)=‌and modulus of denominator
=2‌θ=tan−1(−‌)=−‌‌∴(‌)n‌=(‌)n‌=[cis(‌−(−‌))]n‌=[cis(‌)]n=cis(‌)and
cis(‌)=−1‌=cis(π)‌∴‌‌‌=π‌⇒‌‌n=3thus,
p=3Similarly, second expression
‌(‌)m=cis(‌)⇒[‌]m=cis(‌)⇒[cis(−‌−‌)]m=cis(‌)⇒cis(−‌⋅m)=cis(‌)Therefore,
−‌⋅m=‌⇒−2πm≡2π⋅bmod(6π)⇒−2m≡2‌bmod(6)⇒2m≡−2≡4‌bmod‌6⇒m=2‌bmod‌3∴ Least positive integer
=m=2 thus,
q=2Hence,
√q2+p2=√32+22=√13
© examsiri.com
Go to Question: