Question

Find the modulus and argument of the complex number `(1 + 2"i")/(1 - 3"i")`

Answer 1

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`