Convert of the complex number in the polar form: `sqrt()+i`

Answer

N/a

Explanation

Let `sqrt(3)+ i=r(costheta+ isintheta)`
`therefore" "rcostheta= sqrt(3)`
`" "rsintheta=1`
Squaring and adding the above two equations, we get
`r^(2)cos^(2)theta+ r^(2)sin ^(2)theta=(sqrt(3))^(2) +(1)^(2)`
` rArr" " r^(2)(cos^(2)theta +sin^(2)theta) =3+1=4 `
`rArr" "r^(2) =4" "rArrr=2`
Put `r` = 2 in the above two equations,
`{:(,"",costheta=(sqrt(3))/(2)rArr costheta=30^(@)),(,and, sintheta=(1)/(2)rArrsintheta=sin30^(@)or sin150^(@)):}}`
`rArr" "theta=30^(@)=(pi)/(6)`
then ` " "sqrt(3)+ i=2(cos""(pi)/( 6)+isin""(pi)/(6))`