Convert the complex number `z=(i-1)/(cospi/3+isinpi/3)`in the polar form.

Explanation

Let `z=((i-1))/(("cos"(pi)/(3)+"i sin"(pi)/(3)))=((i-1))/(((1)/(2)+i(sqrt(3))/(2)))=(2(i-1))/((1+ i sqrt(3)))`
`rArr" "z=(2(i-1))/((1+ i sqrt(3)))xx((1-i sqrt(3)))/((1-i sqrt(3)))=((i-1)(1-i sqrt(3)))/(2)`
`" "={((sqrt(3)-1))/(2)+((sqrt(3)+1))/(2)i}`.
Let its polar form be `z = r (cos theta + i sin theta)`.
Now, `r^(2)=|z|^(2)={((sqrt(3)-1)^(2))/(4)+((sqrt(3)+1)^(2))/(4)}=(8)/(4)=2 rArr r = sqrt(2)`.
Let `alpha` the acute angle given by
`tan alpha=|(Im(z))/(Re(z))|=|((sqrt(3)+1)^(2))/(4)xx(2)/((sqrt(3-1)))|=((sqrt(3)+1))/((sqrt(3)-1))=((1+(1)/(sqrt(3))))/((1-(1)/(sqrt(3))))`
`rArr" "tan alpha = (("tan"(pi)/(4)+"tan"(pi)/(6)))/((1-"tan"(pi)/(4)."tan"(pi)/(6)))=tam((pi)/(4)+(pi)/(4))="tan"(5pi)/(12)`
`rArr" "alpha=(5pi)/(12)`.
Clearly, the point representing the given complex number is `P((sqrt(3)-1)/(2),(sqrt(3)+1)/(2))`, which lies in the first quadrant.
`therefore" "arg(z) = theta = alpha = (5pi)/(12) rArr theta = (5pi)/(12)`.
Thus, `r = |z| = theta = alpha = (5pi)/(12) rArr theta = (5pi)/(12)`.
Hence, the required polar form is `z = sqrt(2)("cos"(5pi)/(12)+"i sin"(5pi)/(12))`.