(1+i)^(-3)...

(1+i)^(-3)

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(-2-(1)/(3)i)^(3)

What is modulus of (1)/(1+3i)-(1)/(1-3i) ?

For i^2= -1, (1 - 3i)^3= ?

(1+3i)(1-3i)

((1)/(3)+3i)^(3)=((1)/(3))^(3)+(3i)^(3)+3(1)/(3))(( 1)/(3)+3i)

abs(1/((3 +i)^2)-1/((3-i)^2)) =

" If "z=((1+i)(1+2i)(1+3i))/((1-i)(2-i)(3-i))" then the principal argument of "z"

the value of ((1+i sqrt(3))/(1-i sqrt(3)))^(6)+((1-i sqrt(3))/(1+i sqrt(3)))^(6) is

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