How do you find the 3rd root of #8e^(45i)#?

1 Answer
Aug 24, 2016

Roots are #2[cos (pi/12) +isin(pi/12)], 2[cos ((3pi)/4) +((isin(3pi))/4)] & 2[cos ((17pi)/12) +((isin(17pi))/12)]#

Explanation:

#8e^(45i)=8(cos45+isin45)=8(cos (pi/4)+isin(pi/4)):.# 1st root: #[8(cos pi/4)+(isinpi/4)]^(1/3)= 8^(1/3)[ cos (pi/(4*3)) +isin(pi/(4*3)]=2[cos (pi/12) +isin(pi/12)] :.#for 2nd root #(2pi)/3# to be added to the angle i.e #theta=(pi/12+(2pi)/3)=(9pi)/12=(3pi)/4:.#2nd root is #2[cos ((3pi)/4) +((isin(3pi))/4)]#
for 3rd root #(2pi)/3# to be added to the angle i.e #theta=((3pi)/4+(2pi)/3)=(17pi)/12:.#3rd root is #2[cos ((17pi)/12) +((isin(17pi))/12)]#[Ans]