सम्मिश्र संख्या `z = (i-1)/("cos"(pi)/(3) + i "sin"(pi)/(3))` धुवीय रूप में परिवर्तित कीजिए।

Explanation

`z = (i-1)/("cos"(pi)/(3) + i"sin"(pi)/(3))`
`= (i-1)/((1)/(2) + (sqrt3)/(2)i)`
`= (2(i-1))/(1 + sqrt3i)`
`= (2(i-1))/(1 + sqrt3i) xx (1 - sqrt3i)/(1 - sqrt3i)`
`= (2(i + sqrt3 -1 + sqrt3i))/(1 + 3)`
`= (sqrt3-1)/(2) + (sqrt3 + 1)/(2)i = r (cos theta + isin theta)`
`r cos theta= (sqrt3-1)/(2), r sin theta = (sqrt3 + 1)/(2)`
वर्ग करके जोड़ने पर
`r^(2) = ((sqrt3-1)/(2))^(2) +((sqrt3 + 1)/(2))^(2) = 2`
या `r = sqrt2`
`"cos" theta = (sqrt3 -1)/(2 sqrt2), sin theta = (sqrt3 +1)/(2sqrt2)`
`:' tan theta = |((sqrt3 + 1)/(2 sqrt2))/((sqrt3 -1)/(2 sqrt2))| = (sqrt3 +1)/(sqrt3-1)`
`= (1 + (1)/(sqrt3))/(1 - (1)/(sqrt3)) = ("tan"(pi)/(4) - "tan"(pi)/(6))/(1-"tan"(pi)/(4)."tan"(pi)/(6))`
`= tan ((pi)/(4) + (pi)/(6))`
`:. theta = (pi)/(4) + (pi)/(6) = (5pi)/(12)` अर्थात धुवीय रूप `sqrt2 ("cos"(5pi)/(12) + i "sin" (5pi)/(12))`