Question

convert the complex number 3(cospi/3 -isinpi/6) into polar form

Answer 1

Z= (i-1)/[cos(π/3) + i.sin(π/3)]
Z= (i-1)/[(1/2) + i.(√3)/2]
Z= (i-1)/[(1+√3.i)/2]
Z= 2(i-1)/[1+√3.i]
Z= [(2i-2)*(1-√3.i)]/[(1+√3.i)*(1- √3.i)]
Z= [2i - 2√3(i^2) - 2 + 2√3.i]/[1 - 3(i^2)]
Z= [(-2 + 2√3) + (2√3 + 2).i]/(1+ 3)
Z= [(-2 + 2√3) + (2√3 + 2).i]/4
Z= (-2 + 2√3)/4 + ((2√3 + 2).i)/4
Z= (-1 + √3)/2 + [(√3 + 1)/2]i

Thus, x=(√3-1)/2, y=(√3+1)/2

|Z| = r = √{[(√3-1)/2]^2 + [(√3-1)/2]^2}
     = √{[(√3-1)^2]/4 + [(√3+1)^2]/4}
     = √{[(3-2√3+1)+(3+2√3+1)]/4}
     = √{[4+4]/4}
     = √{8/4}
     = √2

Now,
  As x > 0 & y > 0
Theta(the symbol) = tan^(-1)[y/x] ...... (tan inverse of y by x)
= tan^(-1){[(√3+1)/2][(√3-1)/2]}
= tan^(-1)[(√3+1)/(√3-1)]
= tan^(-1)[(√3+1)*(√3+1)/(√3-1)*(√3+1)]
= tan^(-1)[(3+2√3+1)/(3-1)]
= tan^(-1)[(4+2√3)/2]
= tan^(-1)[2+√3]


Therefore,
the polar form is,
Z = r[cos(theta) + i sin(theta)]
Z = √2[cos(theta) + i sin(theta)]
where theta = tan^(-1)[2+√3]

pls add my answer to the brain list answer.