For a non-zero complex number z , let arg ( z ) denote the principal argument of z , with arg ( z ) . Let be the cube root of unity for which 0 arg ( . Let = arg ( n = 1 2025 ( -) n ) . Then the value of 3 is _________. [2025]

Question

For a non-zero complex number $z$ , let $\arg (z)$ denote the principal argument of $z$ , with $- π < \arg (z) \leq π$ . Let $ω$ be the cube root of unity for which $0 < \arg (ω) < π$ . Let $α = \arg (\sum_{n = 1}^{2025} {(- ω)}^{n}) .$ Then the value of $\frac{3 α}{π}$ is _________. [2025]

Smart Achievers · Accepted Answer

(-2)

$α = \arg (- ω + ω^{2} - ω^{3} + \dots + {(- ω)}^{2025})$

$α = \arg (\frac{- ω ({(- ω)}^{2025} - 1)}{- ω - 1})$ $\frac{3 α}{π}$

$\Rightarrow α = \frac{- 2 π}{3} \Rightarrow \frac{3 α}{π} = - 2$

Solution

Ans.
(-2)

$α = \arg (- ω + ω^{2} - ω^{3} + \dots + {(- ω)}^{2025})$

$α = \arg (\frac{- ω ({(- ω)}^{2025} - 1)}{- ω - 1})$ $\Rightarrow α = \arg (2 ω^{2})$

$\Rightarrow α = \frac{- 2 π}{3} \Rightarrow \frac{3 α}{π} = - 2$

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Solution

Q. For a non-zero complex number z, let arg(z) denote the principal argument of z, with -π<arg(z)≤π. Let ω be the cube root of unity for which 0<arg(ω)<π. Let α=arg(∑n=12025(-ω)n). Then the value of 3απ is _________. [2025]

Ans. (-2) α=arg(-ω+ω2-ω3+⋯+(-ω)2025) α=arg(-ω((-ω)2025-1)-ω-1)⇒α=arg(2ω2) ⇒α=-2π3 ⇒ 3απ=-2

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Ans.
(-2)

$α = \arg (- ω + ω^{2} - ω^{3} + \dots + {(- ω)}^{2025})$

$α = \arg (\frac{- ω ({(- ω)}^{2025} - 1)}{- ω - 1})$ $\Rightarrow α = \arg (2 ω^{2})$

$\Rightarrow α = \frac{- 2 π}{3} \Rightarrow \frac{3 α}{π} = - 2$