What is the value of #(-sqrt(-1))^(4n+3)#?

Question

18667 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-22T16:25:29

#(-sqrt(-1))^(4n+3)=i# or #sqrt(-1)#

#(-sqrt(-1))^(4n+3)#

= #(-i)^(4n+3)#, where #i^2=-1#

= #(-1xxi)^(4n+3)#

= #(-1)^(4n+3)xxi^(4n+3)#

= #-1xxi^(4n)xxi^3#, as odd power of #-1# is #-1#

= #-1xx1xxi^2xxi#, as #i^2=-1# #i^4=(i^2)^2=(-1)^2=1#

= #-1xx1xx(-1)xxi#

= #i# or #sqrt(-1)#