If a1-ib1 a2-ib2 - an-ibn -A-iB- then tan-1b1a1-tan-1b2a2-tan-1bnan-

The correct option is B kπ+tan^ 1BA In the r θ Euler form ⇒ (r_1e^iθ _1)(r_2e^iθ _2)...(r_ne^iθ _n)=A+iB ⇒ (r_1r_2...r_n)(e^i(θ _1+θ _2...θ ...

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Complex Numbers

If a1+ib1 ...

Question

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

k π

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

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k π + {tan}^{- 1} \frac{B}{A}

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k π - {tan}^{- 1} \frac{B}{A}

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{tan}^{- 1} [20 \sum k = 0 sec (\frac{5 π}{12} + \frac{k π}{2}) sec (\frac{5 π}{12} + \frac{(k + 1) π}{2})] = - {tan}^{- 1} a

a

0 < a_{1} < a_{2} < . . . . < a_{n}

t a n^{- 1} (\frac{a_{1} x - y}{x + a_{1} y}) + t a n^{- 1} (\frac{a_{2} - a_{1}}{1 + a_{2} a_{1}}) + t a n^{- 1} (\frac{a_{3} - a_{2}}{1 + a_{3} a_{2}}) + . . . + t a n^{- 1} (\frac{a_{n} - a_{n - 1}}{1 + a_{n} a_{n - 1}}) + t a n^{- 1} (\frac{1}{a_{n}})

k_{1} π + t a n^{- 1} \frac{k_{2} x}{y}

8 k_{1} k_{2}

a_{1}, a_{2}, a_{3} . . . ., a_{n}

t a n [t a n^{- 1} (\frac{d}{1 + a_{1} a_{2}})] + t a m^{- 1} (\frac{d}{1 + a_{2} a_{3}})] + . . . + t a n^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] =

a_{1}, a_{2}, a_{3} \dots \dots . . a_{n}

t a n [t a n^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + t a n^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + \dots \dots \dots \dots \dots \dots + t a n^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] =

θ = t a n^{- 1} \frac{d}{1 + a_{1} a_{2}} + t a n^{- 1} \frac{d}{1 + a_{2} a_{3}} + \dots + t a n^{- 1} \frac{d}{1 + a_{n - 1} a_{n}}

a_{1}, a_{2}, a_{3}, \dots a_{n}

t a n θ =

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