Question

Convert the complex number z = `("i" - 1)/(cos  pi/3 + "i" sin  pi/3)` in the polar form

Answer 1

z = `("i" - 1)/(cos  pi/3 + "i" sin  pi/3)`

= `("i" - 1)/(1/2 + "i"(sqrt(3)/2))`

= `(2"i" - 2)/(1 + sqrt(3)"i")`

= `(2"i" - 2)/(1 + sqrt(3)"i") xx (1 - sqrt(3)"i")/(1 - sqrt(3)"i")`

= `(2"i" - 2sqrt(3)"i"^2 - 2 + 2sqrt(3)"i")/(1 - 3"i"^2)`

= `(2"i" + 2sqrt(3) - 2 + 2sqrt(3)"i")/(1 + 3)`   ...[∵ i² = – 1]

= `((-2 + 2sqrt(3)) + (2 + 2sqrt(3))"i")/4`

∴ z = `((-1 + sqrt(3))/2) + ((1 + sqrt(3))/2)"i"`

This is of the form a + bi, where

a = `(-1 + sqrt(3))/2` and b = `(1 + sqrt(3))/2`

∴ r = `sqrt("a"^2 + "b"^2)`

= `sqrt(((sqrt(3) - 1)/2)^2 + ((sqrt(3) + 1)/2)^2`

= `sqrt((3 + 1 - 2sqrt(3))/4 + (3 + 1 + 2sqrt(3))/4)`

= `sqrt(8/4)`

= `sqrt(2)`

Also, cos θ = `"a"/"r" = (sqrt(3) - 1)/(2sqrt(2))`

and sin θ  = `"b"/"r" -= (sqrt(3) + 1)/(2sqrt(2))`

∴ tan θ =`(sqrt(3) + 1)/(sqrt(3) - 1)`

∴ the polar form of z = r(cos θ + i sin θ)

= `sqrt(2)(cos theta + "i" sin theta)`,

where tan θ = `(sqrt(3) + 1)/(sqrt(3) - 1)`