`sqrt(3)+i`, जहाँ `i=sqrt(-1)` है, का मापांक-आयाम रूप है

Options

A. `2(cos.(pi)/(3)+ isin.(pi)/(3))`
B. `2(cos.(pi)/(6)+ isin.(pi)/(6))`
C. `4(cos.(pi)/(3)+ isin.(pi)/(3))`
D. `4(cos.(pi)/(6)+ isin.(pi)/(6))`

Answer

B

Explanation

माना ` z = sqrt3 + i`
` :. |z| = sqrt((sqrt3)^(2)+(1)^(2))`
` [:' z = a + ib rArr |z| = sqrt(a^(2)+b^(2))]`
` = sqrt(3+1)`
` = sqrt4 = 2`
अब , `arg (z) = tan^(-1)(b/a)`
` = tan^(-1)(1/sqrt3)`
` = pi/6" "[:' tan pi//6 = 1/sqrt3]`
` :. z = r(cos theta + i sin theta)`
` = 2(cos pi//6 + i sin pi //6)`
` [r = |z|= 2 "और " theta = arg (z) = pi/6]`