ध्रुवीय रूप में लिखिए
` z = ( i - 1 ) /( cos "" (pi ) /( 3 ) + i sin "" ( pi ) /( 3 )) `

Explanation

` z = - ( 1 - i ) /( ( 1 )/ ( 2 ) + i (sqrt3 ) / ( 2 ) ) = - ( 2 ( 1 - i )) /( 1 + i sqrt 3 ) `
` = ( - 2 ( 1 - i ) (1 - i sqrt 3)) /( (1 - i sqrt 3 ) ( 1 - i sqrt 3 )) `
` = ( - 2 [ 1- sqrt 3 - ( 1 + sqrt 3 ) i ] ) /( 1 + 3 ) `
तो ` a = ( sqrt3 - 1 ) /( 2 ) = b = (sqrt 3 + 1 ) /( 2 ) (sqrt 3 - 1 ) /( 2) + ( sqrt 3 + 1) / ( 2 ) i = a + ib ` (माना )
अब ` |z|^ 2 = a ^ 2 + b ^ 2 = ((sqrt3 - 1 ))/ ( 4 ) + ((sqrt3 + 1 ) ^ 2 ) /( 4 ) = 2 `
` therefore r = |z| = sqrt 2 `
तथा ` tan theta = | ( b) / ( a ) | = | (sqrt 3 + 1 ) /( sqrt 3 - 1) | `
` = ( sqrt3 + 1 ) / ( sqrt3 - 1 ) = ( 1 + ( 1 ) /( sqrt 3)) /( 1 - ( 1 ) /( sqrt 3 )) = tan (( pi ) /( 4 ) + ( pi ) / ( 6 )) `
= ` tan "" ( 5pi ) /( 12 ) `
` therefore " " theta = ( 5pi ) / ( 12 ) `
यहाँ ` a gt 0 , b gt 0 therefore a + ib ` प्रथम वाद में होगा
अतः `arg = ( 5pi ) / ( 12 ) `
z का ध्रुवीय रूप ` = sqrt 2 (cos "" ( 5pi ) /( 12 ) + isin "" ( 5pi ) /( 12 ) ) `