Let arg(zk)=n(2k+1)π where k=1,2,………n. If arg(z1,z2,z3,………….zn)=π, then n must be of form (m∈z)

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Let {a}{r}{g{{({z}_{{{k}}})}}}={{{({2}{k}+{1})}}}{{{n}}} where

Question

Let $a r g (z_{k}) = \frac{( 2 k + 1 ) π}{n}$ where $k = 1, 2, \dots\dots\dots n$ . If $a r g (z_{1}, z_{2}, z_{3}, \dots\dots\dots\dots . z_{n}) = π$ , then $n$ must be of form $(m \in z)$

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Benjamin Discussed

Let

a r g (z_{k}) = \frac{( 2 k + 1 ) π}{n}

where

k = 1, 2, \dots\dots\dots n

. If

a r g (z_{1}, z_{2}, z_{3}, \dots\dots\dots\dots . z_{n}) = π

, then

n

must be of form

(m \in z)

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Question Text

Let

a r g (z_{k}) = \frac{( 2 k + 1 ) π}{n}

where

k = 1, 2, \dots\dots\dots n

. If

a r g (z_{1}, z_{2}, z_{3}, \dots\dots\dots\dots . z_{n}) = π

, then

n

must be of form

(m \in z)