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Value of z satisfying the equation $\log z+\log z^2+.......+\log z^n=0$ is

$\begin{array}{1 1}(A)\;\sin\large\frac{4m\pi}{n}\normalsize +i\cos\large\frac{4m\pi}{n}\normalsize ,m=1,2....\\(B)\;\cos\large\frac{4m\pi}{n(n+1)}\normalsize -i\sin\large\frac{4m\pi}{n(n+1)}\normalsize ,m=1,2....\\(C)\;\cos\large\frac{4m\pi}{n(n+1)}\normalsize +i\sin\large\frac{4m\pi}{n(n+1)}\normalsize ,m=1,2....\\(D)\;0\end{array} $

1 Answer

$\log z+\log z^2+......+\log z^n=\log z,z^2.......z^n$
$\Rightarrow \log z^{\Large\frac{n(n+1)}{2}}=0$
$\Rightarrow z^{\Large\frac{n(n+1)}{2}}=1$
$(\large\frac{n(n+1)}{2})^{th}$ root of unity=$\cos\large\frac{4m\pi}{n(n+1)}\normalsize +i\sin\large\frac{4m\pi}{n(n+1)}\normalsize ,m=1,2....$
Hence (C) is the correct answer.
answered Apr 10, 2014 by sreemathi.v
 
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