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 $i ^18 = -1$

How do you simplify i^18i^18?