How do you verify that De Moivre's Theorem holds for the power n=0?

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Answer 1 · 2017-03-28T01:00:12

De Moivre's theorem tells us that;

# (cos theta + isin theta)^n = cos n theta + isin n theta #

We can prove the case #n=0# directly from Euler's Formula:

# e^(ix) = cos x + isin x \ \ \ AA x in RR #

Consider the #LHS# with #n=0#:

# (cos theta + isin theta)^n = (cos theta + isin theta)^0 #
# " " = (e^(i theta))^0 \ \ # (using Euler's Formula)
# " " = e^0 #
# " " = 1 #

Consider the #RHS# with #n=0#

# cos n theta + isin n theta = cos 0 + isin 0 #
# " " = 1 + 0 #
# " " = 1 #

And so with #n=0# we have #LHS=RHS# QED