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Complex Numbers

If a+ib 5=α...

Question

If ${(a + i b)}^{5} = α + i β$ , then ${(b + i a)}^{5}$ is equal to

A

β + i α

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B

α - i β

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C

β - i α

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D

- α - i β

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Solution

The correct option is A $β + i α$
$(a + i b)^{5} = a^{5} + 5 a^{4} b i - 10 a^{3} b^{2} - 10 a^{2} b^{3} i + 5 a b^{4} + i b^{5} = α + i β$

$(a - i b)^{5} = a^{5} - 5 a^{4} b i - 10 a^{3} b^{2} + 10 a^{2} b^{3} i + 5 a b^{4} - i b^{5} = α - i β$

$(b + i a)^{5} = i^{5} (- b i + a) = i (a - b i)^{5}$ $= i (α - i β) = β + i α$

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Similar questions

α

β

∣ ∣ ∣ \frac{α + i β}{β + i α} ∣ ∣ ∣ =

(a + i b)^{5} = α + i β

(b + i a)^{5}

Q. If the roots of the equation

x^{4} - 6 x^{3} + 18 x^{2} - 30 x + 25 = 0

are of the form

α \pm i β

β \pm i α

(α, β) =

(a + i b)^{5} = α + i β t h e n (b + i a)^{5}

e^{x + i y} = α + i β,

x + i y

is called logarithm of

α + i β

∴ {log}_{e} (x + i y) = {log}_{e} (r e^{i θ})

= {log}_{e} r + i θ

where r is modulus value of

x + i y

θ

be the argument of

x + i y

i^{α + i β)} = α + i β,

α^{2} + β^{2}

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