Verify using polar form that #i^9# is #i# Can someone please help me with this? I tried doing this and came up with z=r=#sqrt1#(cos90 + #i#sin90) And #z^9#=#sqrt1#^9(cos 810 + #i# sin 810)
1 Answer
Mar 26, 2017
See calculation below
Explanation:
Applying De moivre's Theorem
So,
You were on the good path, you needed only to simplify
I hope this is helpful