1.

Polar form of sqrt(3) + i is

Answer»

`-2(cos""(PI)/( 6)+isin""(pi)/(6))`
`2(cos""(pi)/( 6)+isin""(pi)/(6))`
`-2(cos""(2pi)/( 3)+isin""(pi)/(6))`
`2(cos""(2pi)/( 3)+isin""(pi)/(6))`

Solution :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))`


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