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