Question

Answer the following:

Convert the complex numbers in polar form and also in exponential form.

z = `-6 + sqrt(2)"i"`

Answer 1

z = `-6 + sqrt(2)"i"`

∴ a = – 6, b = `sqrt(2)`, i.e. a < 0, b > 0

∴ r = `sqrt("a"^2 + "b"^2)`

= `sqrt((-6)^2 + (sqrt(2))^2`

= `sqrt(36 + 2)`

= `sqrt(38)`

Here `(-6, sqrt(2))` lies in 2^nd quadrant

∴ amp (z) = θ 

= `pi + tan^-1("b"/"a")`

= `tan^-1(-sqrt(2)/6) + pi`

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

∴ `sqrt(38)(cos theta + "i" sin theta)`, where θ

= `pi + tan^-1(-sqrt(2)/6)`

∴ The exponential form of z = re^iθ

`sqrt(38)"e" ^(pi + tan^-1(-sqrt(2)/6)`