Question

If $nϵN>1$, 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: (D)

$\frac{(1-n)}{2}$

$|\frac{z+1}{z}{|}^n=|1|$
$|z+1{|}^n=|z{|}^n$
Let $z=x+iy$
$(x+1{)}^2+y^2=(x{)}^2+y^2$
$(2x+1)\left(1\right)=0$
$2x+1=0$
$x=\frac{-1}{2}$
Thus $Re\left(z\right)=\frac{-1}{2}$
Now the roots will be collinear and will have their real parts as $\frac{-1}{2}$
For the equation
$(z+1{)}^n=z^n$
There will be exactly $n-1$ roots.
Hence sum of the real parts of all the roots of the equation will be
$=(n-1)\left(\frac{-1}{2}\right)$
$=\frac{1-n}{2}$