Question

If (a1+ib1)(a2+ib2)....(an+ibn)= A+iB, then prove Tan-1(b1/a1)+ Tan-1(b2/a2)+...+ Tan-1=nπ+tan^-1(B/A)​

Answer 1

Answer:

$tan^{-1}{\frac{b_1}{a_1}}+tan^{-1}{\frac{b_2}{a_2}}+....+tan^{-1}{\frac{b_n}{a_n}}=n\pi+tan^{-1}{\frac{b_n}{a_n}}$

Step-by-step explanation:

Formula used:

The argument of the complex number z=x+iy is $arz=tan^{-1}{\frac{y}{x}$

Given:

$(a_1+ib_1)(a_2+ib_2)....(a_n+ib_n)= A+iB$

Taking arguement on both sides, we get

$arg[(a_1+ib_1)(a_2+ib_2)....(a_n+ib_n)]= arg[A+iB]$

$\implies\:arg(a_1+ib_1)+arg(a_2+ib_2)+....+arg(a_n+ib_n)]= arg[A+iB]$

$\implies\:tan^{-1}{\frac{b_1}{a_1}}+tan^{-1}{\frac{b_2}{a_2}}+....+tan^{-1}{\frac{b_n}{a_n}}=tan^{-1}{\frac{b_n}{a_n}}$

Taking general value

$\implies\:tan^{-1}{\frac{b_1}{a_1}}+tan^{-1}{\frac{b_2}{a_2}}+....+tan^{-1}{\frac{b_n}{a_n}}=n\pi+tan^{-1}{\frac{b_n}{a_n}}$