How do you simplify #i^18#?

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Answer 1 · 2015-11-14T10:11:53

It is #i^18=(i^2)^9=(-1)^9=-1#

Assuming that #i# is the imaginary unit with #i^2=-1#

Answer 2 · 2015-11-14T10:23:05

#i ^18 = -1#

Consider the following:

#a^2 times a^3 = a times a times a times a times a= a^5 = a^(2+3)#

Now look at your question:

#i^18 = i ^(2 + 16) = i^(2) times i^(16#

#i =sqrt(-1) " so " i ^2 =-1#

#i^16 # is 8 lots of #i^2# multiplied together.

An even number of negatives multiplied together yields a positive number so:

#i ^16 = +1#

But we have

#i ^2 times i ^16 = (-1) times (+1) = -1#

So #i ^18 = -1#