Question

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

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

Answer 1

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

We have,
$\log z+\log z^2+...+\log z^n=0$

$\Rightarrow \log (z\cdot z^2\cdots z^n)=0$

$\Rightarrow \log z^{\left\{\frac{n(n+1)}{2}\right\}}=0$

$\Rightarrow z^{\frac{n(n+1)}{2}}=1$

$\Rightarrow z^{\frac{n(n+1)}{2}}=e^{2m\pi },m=0,1,2,\dots$

$\Rightarrow z=e^{\frac{4m\pi }{n(n+1)}}$

$\Rightarrow z=\cos \frac{4m\pi }{n(n+1)}+i\sin \frac{4m\pi }{n(n+1)},m=0,1,2,\dots$

Hence, option (A) is correct.