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प्रश्न
Answer the following:
Convert the complex numbers in polar form and also in exponential form.
z = `(2 + 6sqrt(3)"i")/(5 + sqrt(3)"i")`
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उत्तर
z = `(2 + 6sqrt(3)"i")/(5 + sqrt(3)"i")`
= `(2 + 6sqrt(3)"i")/(5 + sqrt(3)"i") xx (5 - sqrt(3)"i")/(5 - sqrt(3)"i")`
= `(10 - 2sqrt(3)"i" + 30sqrt(3)"i" - 18"i"^2)/(25 - 3"i"^2)`
= `(10 + 28sqrt(3)"i" + 18)/(25 + 3)` ...[∵ i2 = – 1]
= `(28 + 28sqrt(3)"i")/28`
∴ z = `1 + sqrt(3)"i"`
This is of the form a + bi, where a = 1, b = `sqrt(3)`
∴ r = `sqrt("a"^2 + "b"^2)`
= `sqrt(1^2 + (sqrt(3))^2`
= `sqrt(1 + 3)`
= 2
If θ is the amplitude, then cos θ = `"a"/"r" = 1/2`
and sin θ = `"b"/"r" = sqrt(3)/2`
∴ θ = `pi/3 ...[because cos pi/3 = 1/2 and sin pi/3 = sqrt(3)/2]`
∴ polar form of z = r(cos θ + i sin θ)
= `2(cos pi/3 + "i" sin pi/3)`
and the exponential form of z = reiθ
= `2"e"^("i"(pi/3))`
= `2"e"^(pi/3"i")`
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