Let p and q be positive integers having no positive common divisors except unity. Let z_1 , z_2 ,..., z_q be the q values of z^(p/q) , where z is a fixed complex number. Then the product z_1 z_2 ...z_q is equal to (A) z^p , if q is odd (B) −z^p , if q is even (C) z^p , if q is even (D) −z^p , if q is odd


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Question Number 140248 by EnterUsername last updated on 05/May/21

$Let p and q be positive integers having no positive$ $common divisors except unity . Let z_{1}, z_{2}, . . ., z_{q} be the$ $q values of z^{p / q}, where z is a fixed complex number . Then$ $the product z_{1} z_{2} . . . z_{q} is equal to$ $(A) z^{p}, if q is odd (B) - z^{p}, if q is even$ $(C) z^{p}, if q is even (D) - z^{p}, if q is odd$
	Answered by mr W last updated on 05/May/21
	$z=re^(θi) =re^((2kπ+θ)i) z_k =z^(p/q) =r^(p/q) e^((((2kπ+θ)p)/q)i) ,k=0,1,...,q−1 Πz_k =r^p e^((((2π×((q(q−1))/2)+qθ)p)/q)i) =r^p e^((qπ−π+θ)pi) =(re^((qπ−π+θ)i) )^p =(re^(θi) e^((q−1)πi) )^p =(ze^((q−1)πi) )^p =z^p e^(p(q−1)πi) = { ((z^p if p(q−1) is even)),((−z^p if p(q−1) is odd)) :} (A) if q is old, p(q−1) is even, result is z^p ⇒ correct (B) if q is even, then p is odd, then p(q−1) is odd, the result is −z^p , ⇒correct.$
	$z = {r e}^{θ i} = {r e}^{(2 k π + θ) i}$ $z_{k} = z^{p / q} = r^{p / q} e^{\frac{(2 k π + θ) p}{q} i}, k = 0, 1, . . ., q - 1$ $Π z_{k} = r^{p} e^{\frac{(2 π \times \frac{q (q - 1)}{2} + q θ) p}{q} i} = r^{p} e^{(q π - π + θ) p i}$ $= {({r e}^{(q π - π + θ) i})}^{p}$ $= {({r e}^{θ i} e^{(q - 1) π i})}^{p}$ $= {({z e}^{(q - 1) π i})}^{p}$ $= z^{p} e^{p (q - 1) π i}$ $= {\begin{cases} z^{p} i f p (q - 1) i s e v e n \\ - z^{p} i f p (q - 1) i s o d d \end{cases}$ $(A) i f q i s o l d, p (q - 1) i s e v e n,$ $r e s u l t i s z^{p} \Rightarrow c o r r e c t$ $(B) i f q i s e v e n, t h e n p i s o d d, t h e n$ $p (q - 1) i s o d d, t h e r e s u l t i s - z^{p},$ $\Rightarrow c o r r e c t .$
	Commented by EnterUsername last updated on 05/May/21

	$Thank you, Sir$
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