Let $z_1=2[cos(pi/6)+isin(pi/6)]$ How do you find $1/z_1?$

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Answer 1 · 2016-12-05T09:25:28

$1/z_1=sqrt3/4-i1/4$

$1/(z_1)=z_1^-1$

using De'Moivre's theorem

$1/z_1=z_1^-1=[2(cos(pi/6)+isin(pi/6))]^-1$

$1/z_1=z_1^-1=[2^-1(cos(-pi/6)+isin(-pi/6))]$

$costheta$ is an even function; $""sintheta$ an odd one#

$1/z_1=z_1^-1=[1/2(cos(pi/6)-isin(pi/6))]$

$1/z_1=z_1^-1=[1/2(sqrt3/2-i1/2)]$

$1/z_1=sqrt3/4-i1/4$

Let z_1=2[cos(pi/6)+isin(pi/6)]z_1=2[cos(pi/6)+isin(pi/6)] How do you find 1/z_1? 1/z_1?