Question

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

• A. $cos\frac{4m\pi }{n(n+1)}+isin\frac{4m\pi }{n(n+1)},m=1,2,...$
• B. $cos\frac{4m\pi }{n(n+1)}-isin\frac{4m\pi }{n(n+1)},m=1,2,...$
• C. $sin\frac{4m\pi }{n}+icos\frac{4m\pi }{n},m=1,2,...$
• D. $0$

Answer 1

The correct option is A $cos\frac{4m\pi }{n(n+1)}+isin\frac{4m\pi }{n(n+1)},m=1,2,...$

$log\left(z\right)+log\left(z^2\right)+...log\left(z^n\right)=0$
Let $z=e^{i\theta }$
Hence
$log\left(z\right)+2log\left(z\right)+3log\left(z\right)...+nlog\left(z\right)=0$
$log\left(z\right)(1+2+3+...+n)=0$
$log\left(z\right)(\frac{n(n+1)}{2})=0$
$log\left(z\right)(\frac{n(n+1)}{2})=0$
$log\left(z^{\frac{n(n+1)}{2}}\right)=log\left(1\right)$
$z^{\frac{n(n+1)}{2}}=1$
$e^{i\theta \frac{n(n+1)}{2}}=e^{2im\pi }$
$\theta \frac{n(n+1)}{2}=2m\pi$
$\theta =\frac{4m\pi }{n(n+1)}$
$z=e^{i\theta }$
$=e^{i\frac{4m\pi }{n(n+1)}}$
$=cos(\frac{4m\pi }{n(n+1)})+isin(\frac{4m\pi }{n(n+1)})$