If (a1+ib1)(a2+ib2).....(an+ibn)=A+iB then Tan−1(b1a1)+Tan−1(b2a2)+.....Tan−1(bnan)=______

Arg (a1+ib1)+Arg(a2+ib2)+.......+Arg(an+ibn)=Arg (A+iB) ⇒Tan−1(b1a1)+Tan−1(b2a2)+.........+Tan−1(bnan)=Tan−1(BA) =2nπ+Tan−1(BA)

If (a1+ib1)(a2+ib2).....(an+ibn)=A+iB then Tan−1(b1a1)+Tan−1(b2a2)+.....Tan−1(bnan)=______

Arg (a1+ib1)+Arg(a2+ib2)+.......+Arg(an+ibn)=Arg (A+iB) ⇒Tan−1(b1a1)+Tan−1(b2a2)+.........+Tan−1(bnan)=Tan−1(BA) =2nπ+Tan−1(BA)

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If $(a_{1} + i b_{1}) (a_{2} + i b_{2}) ..... (a_{n} + i b_{n}) = A + i B$ then $T a n^{- 1} (\frac{b_{1}}{a_{1}}) + T a n^{- 1} (\frac{b_{2}}{a_{2}}) + ..... T a n^{- 1} (\frac{b_{n}}{a_{n}}) =______$

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$T a n^{- 1} (\frac{A}{B})$

$2 n π + T a n^{- 1} (\frac{B}{A})$

$T a n^{- 1} (\frac{2 A}{B})$

$n π + T a n^{- 1} (\frac{B}{A})$

answer is B.

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Detailed Solution

$A r g (a_{1} + i b_{1}) + A r g (a_{2} + i b_{2}) + ....... + A r g (a_{n} + i b_{n}) = A r g (A + i B) \Rightarrow T a n^{- 1} (\frac{b_{1}}{a_{1}}) + T a n^{- 1} (\frac{b_{2}}{a_{2}}) + ......... + T a n^{- 1} (\frac{b_{n}}{a_{n}}) = T a n^{- 1} (\frac{B}{A}) = 2 n π + T a n^{- 1} (\frac{B}{A})$