If a1 - ib1a2 - ib2-an - ibn - A - iB- then -i - 1n tan - 1 biai is equal to

The correct option is B tan^ 1( B/A) (a_1+ib_1)(a_2+ib_2)...(a_n+ib_n)=A+iB ∵ argument(z)=tan^ 1(ba) arg[(a_1+ib_1)(a_2+ib_2)...(a_n+ib_n)] ...

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Algebra of Derivatives

If a1 + ib1...

Question

If $(a_{1} + i b_{1}) (a_{2} + i b_{2}) . . . (a_{n} + i b_{n}) = A + i B$ , then $n \sum i = 1 {tan}^{- 1} (\frac{b_{i}}{a_{i}})$ is equal to

\frac{B}{A}

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tan (\frac{B}{A})

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{tan}^{- 1} (\frac{B}{A})

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{tan}^{- 1} (\frac{A}{B})

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Solution

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Similar questions

(a_{1} + i b_{1}) (a_{2} + i b_{2}) . . . (a_{n} + i b_{n}) = A + i B

{tan}^{- 1} \frac{b_{1}}{a_{1}} + {tan}^{- 1} \frac{b_{2}}{a_{2}} + . . . {tan}^{- 1} \frac{b_{n}}{a_{n}} =

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(a_{1} + i b_{1}) (a_{2} + i b_{2}) . . . (a_{n} + i b_{n}) = A + i B

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Δ A B C

C

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$t a n^{- 1} a + t a n^{- 1} b$ , where a>0,b>0,ab>1, is equal to

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