If 1 - i 1 - 2i 1 - 3i - 1 - ni - - - i- then 2 - 5 - 10 - 1 - n2 -

The correct option is D α^2+β^2 Given that #160; (1 + i) ( 1 + 2i) ( 1 + 3i) .... (1 + ni) = α + iβ ,If we take modulus on both sides, |a × b| ...

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Modulus of a Complex Number

If 1 + i 1 ...

Question

If $(1 + i) (1 + 2 i) (1 + 3 i) . . . . (1 + n i) = α + i β$ then $2.5.10 . . . (1 + n^{2}) =$

α - i β

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α^{2} - β^{2}

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α^{2} + β^{2}

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none of these

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Solution

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Similar questions

(1 + i) (1 - 2 i) (1 - 3 i) . . . (1 - n i) = α - i β

α^{2} + β^{2}

(1 - i) (1 + 2 i) (1 + 3 i) . . . . (1 + n i) = α + i β

2 \cdot 5 \cdot 10 . . . . (1 + n^{2}) =

z_{0} = α + i β, i = \sqrt{- 1}

x^{3} - 2 (1 + α) x^{2} + (4 α + α^{2} + β^{2}) x - 2 (α^{2} + β^{2}) = 0

e^{x + i y} = α + i β,

x + i y

α + i β

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

= {log}_{e} r + i θ

x + i y

θ

x + i y

i^{α + i β)} = α + i β,

α^{2} + β^{2}

(1 + i) (1 + 2 i) (1 + 3 i) . . . . (1 + n i) = x + i y

2.5.10....... (1 + n^{2}) = x^{2} + y^{2}

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