सम्मिश्र संख्या `((1+i)(2+i))/(3-i)`, जहाँ `i = sqrt(-1)`, का कोणांक क्या है?

Options

A. 0
B. `pi//4`
C. `-pi//4`
D. `pi//2`

Answer

D

Explanation

माना `z = ((1+i)(2+i))/((3-i))`
` = (2+3i + i^(2))/(3-i) = (2+3i -1)/(3-i) [because i^(2) = -1]`
` = (1+3i)/(3-i) xx (3+i)/(3+i) = (3 + 10i + 3i^(2))/(9-i^(2))`
` = (3+10i - 3)/(9+1) [because i^(2) = -1]`
` = (10i)/(10) = i`
`therefore z = 0 + i * 1`
अतः `"arg"(z) = "tan"^(-1)((1)/(0))`
` = "tan"^(-1) (oo) = (pi)/(2)`