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

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

((1+i)/(1-i))^(3)-((1-i)/(1+i))^(3)=x+iy,f in d(x,y)

Reduce ((1)/(1+2i)+(3)/(1-i))((3-2i)/(1+3i)) to the form (a + ib).

Find the inverse of the following functions(i) f(x)=sin^(-1)((x)/(3))x in[-3,3]

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

Range of f(x) =1/(1-2cosx) is(i)[ 1/ 3 , 1 ] (ii) [-1,(1)/(3)] (iii) (-oo,-1)uu[(1)/(3),oo) (iv) [-(1)/(3),1]

if f(x)=|[x-3,2x^2-18,3x^3-81],[x-5,2x^2-50,4x^3-500],[1,2,3]| then f(1)f(3)+f(3)f(5)+f(5)f(1) is equalto

if f(x)=|[x-3,2x^2-18,3x^3-81],[x-5,2x^2-50,4x^3-500],[1,2,3]| then f(1)f(3)+f(3)f(5)+f(5)f(1) is equal to

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