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Question
Answer the following:
Convert the complex numbers in polar form and also in exponential form.
z = `-6 + sqrt(2)"i"`
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Solution
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 2nd 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 = reiθ
`sqrt(38)"e" ^(pi + tan^-1(-sqrt(2)/6)`
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