`(1 + 7i)/((2-i)^(2))` का ध्रुवीय रूप ज्ञात कीजिए ।

Options

A. `r ("cos"theta - i "sin"theta)`
B. `r("sin"theta + i "cos"theta)`
C. `r("cos"theta + i "sin"theta)`
D. `r("sin"theta - i "cos"theta)`

Answer

C

Explanation

माना `z = (1+7i)/((2-i)^(2)) = (1+7i)/(4+i^(2)-4i) = (1+7i)/(4-1-4i)`
`rArr z = ((1+7i)/(3-4i)) ((3+4i)/(3+4i))`
` = (3+4i +21i + 28i^(2))/((3)^(2) - (4i)^(2))`
`=(3+25i - 28)/(9+16) = (-25 + 25i)/(25) = -1 + i`
`rArr z = -1 + I = (-1, 1) = r("cos"theta + i"sin"theta)`