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#