Question

Roots of the equation are $(z+1{)}^5=(z-1{)}^5$ are

• A. $\pm i\tan \left(\frac{\pi }{5}\right),\pm i\tan \left(\frac{2}{5}\right)$
• B. $\pm i\cot \left(\frac{\pi }{5}\right),\pm i\cot \left(\frac{2\pi }{5}\right)$
• C. $\pm i\cot \left(\frac{\pi }{5}\right),\pm i\tan \left(\frac{2\pi }{5}\right)$
• D. none of these

Answer 1

The correct option is B $\pm i\cot \left(\frac{\pi }{5}\right),\pm i\cot \left(\frac{2\pi }{5}\right)$

${(z+1)}^5{=(z-1)}^5$
$⟹{\left(\frac{z+1}{z-1}\right)}^5=1=\cos 0+i\sin 0=cis\left(0\right)$
$⟹\frac{z+1}{z-1}={\left(cis\left(0\right)\right)}^{\frac{1}{5}}=cis\frac{2k\pi }{5}$ ...{De Moivre's Theorem}
Where is $k=0,1,2,3,4$

$⟹z=\frac{cis\frac{2k\pi }{5}+1}{cis\frac{2k\pi }{5}-1}=\frac{\cos \frac{k\pi }{5}[\cos \frac{k\pi }{5}+i\sin \frac{k\pi }{5}]}{\sin \frac{k\pi }{5}[i\cos \frac{k\pi }{5}-\sin \frac{k\pi }{5}]}$

$z=-i\cot \frac{k\pi }{5}$
for$k=0,z$ is not defined.
for $k=1,2,3,4$
$z=-i\cot \frac{\pi }{5},-i\cot \frac{2\pi }{5},-i\cot \frac{3\pi }{5},-i\cot \frac{4\pi }{5}⟹z=\pm i\cot \frac{\pi }{5},\pm i\cot \frac{2\pi }{5}$
Ans: B