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प्रश्न
Find the modulus and argument of the complex number `(1 + 2"i")/(1 - 3"i")`
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उत्तर
Let z = `(1 + 2"i")/(1 - 3"i")`
= `(1 + 2"i")/(1 - 3"i") xx (1 + 3"i")/(1 + 3"i")`
= `(1 + 3"i" + 2"i" + 6"i"^2)/(1 - 9"i"^2)`
= `(1 + 5"i" - 6)/(1 + 9)` ...[∵ i2 = – 1]
∴ z = `(-5 + 5"i")/10`
= `-1/2 + 1/2"i"`
This is of the form a + bi, where a = `-1/2`, b = `1/2`
∴ modulus = r
= `sqrt("a"^2 + "b"^2)`
= `sqrt((-1/2)^2 + (1/2)^2)`
= `sqrt(1/4 + 1/4)`
= `1/sqrt(2)`
If θ is the argument, then
cos θ = `"a"/"r"`
= `((-(1)/2))/((1/sqrt(2)))`
= `-1/sqrt(2)`
and sin θ = `"b"/"r"`
= `((1/2))/((1/sqrt(2))`
= `1/sqrt(2)`
`∴ θ = (3pi)/4 ...[(because cos (3pi)/4 = cos(pi - pi/4) = -cos pi/4=), (-1/sqrt(2) and sin (3pi)/4 = sin(pi - pi/4) = sin pi/4 = 1/sqrt(2))]`
Hence, modulus = `1/sqrt(2)` and argurement = `(3pi)/4`
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