यदि `z = x+ iy = ((1)/(sqrt(2)) - (i)/(sqrt(2)))^(-25)`, जहाँ `i = sqrt(-1)` है, तो `(z-sqrt(2))/(z- isqrt(2))` का मूल आयाम क्या है?

Options

A. `pi`
B. `pi//2`
C. `pi//3`
D. `pi//4`

Answer

A

Explanation

`because z= x + iy = ((1)/(sqrt(2)) - (i)/(sqrt(2)))^(-25)`
` = (e^(-i(pi)/(4)))^(-25) = e^(i(25)/(4)pi)`
` = e^(i(6pi + (pi)/(4))) = e^(i((pi)/(4)))`
` = "cos" (pi)/(4) + "i sin" (pi)/(4) = (1)/(sqrt(2)) + i(1)/(sqrt(2))`
`therefore (z-sqrt(2))/(z-isqrt(2)) = ((1)/(sqrt(2)) + i (1)/(sqrt(2)) -sqrt(2))/((1)/(sqrt(2)) + i((1)/(sqrt(2))) -isqrt(2))`
` = (1+i -2)/(1+i -2i) = (-1 +i)/(1-i) xx (1+i)/(1+i)`
` = (-1+i-i +i^(2))/(1-i^(2))`
` = (-1-1)/(1+1) = (-2)/(2) = -1 +0i`
`alpha = "tan"^(-1) |(0)/(-1)| = "tan"^(-1) (0)`
` = "tan"^(-1)|"tan"0| =0`
`therefore theta` (आयाम) `= pi-0 = pi`