Question

If $n\in N>1$, then the sum of real part of roots of $z^n=(z+1{)}^n$ is equal to

• A. $\frac{n}{2}$
• B. $\frac{(n-1)}{2}$
• C. $-\frac{n}{2}$
• D. $\frac{(1-n)}{2}$

Answer 1

Answer: (C)

$-\frac{n}{2}$

$z^n={(z+1)}^n{\left(\frac{z+1}{z}\right)}^n=1\therefore \frac{z+1}{z}=e^{i\frac{2\pi m}{n}},m=0,1,2,........,n-1\frac{1}{z}=-2{\sin }^2\left(\frac{\pi m}{n}\right)+i2\sin \left(\frac{\pi m}{n}\right)\cos \left(\frac{\pi m}{n}\right)=2i\sin \left(\frac{\pi m}{n}\right)[\cos \left(\frac{\pi m}{n}\right)+i\sin \left(\frac{\pi m}{n}\right)]z=\frac{\cos \left(\frac{\pi m}{n}\right)-i\sin \left(\frac{\pi m}{n}\right)}{2i\sin \left(\frac{\pi m}{n}\right)}=\frac{1}{2}(-1-i\cot \left(\frac{\pi m}{n}\right))$

$\therefore$ Sum of real parts of roots = $n(-\frac{1}{2})=-\frac{n}{2}$ ($\because$ n roots are there)