How do you use demoivre's theorem to simplify #[2(cos((pi)/2)+isin((pi)/2))]^8#?

Question

4205 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-15T16:55:23

#[2(cos(pi/2)+isin(pi/2))]^8=256#

According to de Moivre's Theorem we can calculate any integer power of a complex number given in trigonometric form.

The theorem says that:

If a complex number #z# is given in a form:

#z=|z|(cosvarphi+isinvarphi)#

Then the #n-th# power of #z# is:

#z^n=|z|^n(cosnvarphi+isinnvarphi)#

Using this formula we get:

#[2(cos(pi/2)+isin(pi/2))]^8=2^8[cos((8pi)/2)+isin((8pi)/2)]=#

#=256[cos4pi+isin4pi]=256#