सम्मिश्र संख्या `(1+2i)/(1-3i)` का मापांक और कोणांक हैं

Options

A. `(1)/(sqrt(2)), (3pi)/(4)`
B. `(1)/(sqrt(2)), - (3pi)/(4)`
C. `(1)/(2), (3 pi)/(4)`
D. इनमे से कोई नहीं

Answer

A

Explanation

माना `z = (1+2i)/(1-3i)`
`therefore z = (1+2i)/(1-3i) xx (1+3i)/(1+3i) = (1+3i+2i+6i)/(1^(2) - (3i)^(2)) " " [because (a+b)(a-b) = a^(2) -b^(2)]`
` = (1+5i + 6(-1))/(1-9i^(2)) " " [because i^(2) = -1]`
` = (1+5i -6)/(1+9) = (-5 +5i)/(10) = (-1+i)/(2) rArr z = -(1)/(2) + (1)/(2)i`
`therefore |z| = sqrt((-(1)/(2))^(2) + ((1)/(2))^(2)) " " [because a + ib = sqrt(a^(2) + b^(2))]`
` = sqrt((1)/(4) + (1)/(4)) = sqrt((2)/(4)) = sqrt((1)/(2)) = (1)/(sqrt(2))`
अब, `tan theta = |((1)/(2))/(-(1)/(2))| " " [because theta = "tan"^(-1) |(("lm"(z))/("Re"(z)))|]`
`rArr "tan" theta = 1 = "tan" (pi)/(4) rArr theta = (pi)/(4)`
चूँकि z का वास्तविक भाग ऋणात्मक तथा काल्पनिक भाग धनात्मक है ।
अतः बिंदु द्वितीय चतुर्थांश में स्थित है ।
`therefore "arg"(Z) = pi - theta = pi - (pi)/(4) = (3pi)/(4)`
अतः मापांक `= (1)/(sqrt(2))` तथा कोणांक (Z) = `(3 pi)/(4)`