# The roots of the equation $z^n=(1+z)^n$...

The complex roots of the equation

$$z^{n}=(1+z)^{n}$$

$A.$ are vertices of a regular polygon

$B.$ lie on a circle

$C.$ are collinear

$D.$ none of these

NOTE: $z$ is a complex number.

• Here is a technique. Dec 15, 2013 at 6:58

Clearly, $z\ne0,$

So dividing either sides by $z^n,$ we have $$\left(1+\frac1z\right)^n=1$$

Using de Moivre's formula & Euler's formula, $$1+\frac1z=e^{\frac{2m\pi i}n}=\cos\frac{2m\pi}n+i\sin\frac{2m\pi }n$$ where $m$ is any integer

$$\implies \frac1z=\cos\frac{2m\pi}n+i\sin\frac{2m\pi }n-1=i2\sin\frac{m\pi}n\cos\frac{m\pi}n-2\sin^2\frac{m\pi}n$$ $$=2i\sin \frac{m\pi}n\left(\cos\frac{m\pi}n+i\sin \frac{m\pi}n\right)$$ (using $\cos2A=1-2\sin^2A$)

$$z=\frac1{2i\sin \frac{m\pi}n\left(\cos\frac{m\pi}n+i\sin \frac{m\pi}n\right)}=\frac{{\cos\frac{m\pi}n-i\sin \frac{m\pi}n}}{2i\sin \frac{m\pi}n}=-\frac12-i\frac{\cot\frac{m\pi}n}2$$

• Now what can we conclude from this ? Dec 15, 2013 at 7:02
• @Apurv, so if $z=x+iy,x=-\frac12.$ What can we conclude from here? Dec 15, 2013 at 7:03
• Ok.. So the points lie on x=-1/2. Hence the points are collinear. Thanks !! Dec 15, 2013 at 7:06

An easier method, stolen from comment to an answer from a similar thread.

If $z$ satisfies $z^n=(1+z)^n$, then $|z|^n=|1+z|^n$ and: $$|z|=|z+1|$$

The geometric interpretation is that the distance (in the complex plane) from $0$ to $z$ equals the distance from $-1$ to $z$. The locus of points with the same distance from $0$ and $-1$ is the vertical line: $$\mathrm{Re}\ z = -\frac12$$ so every solution to the original equation lies on this line.

Rewrite as $$\left(\frac z{1+z}\right)^n = 1$$. Now from this you can work out the value of $$\frac z{1+z}$$, and note in particular that they lie on the unit circle. And $$\frac z{1+z}$$ is a Möbius transform, so you should be able to figure out how its inverse maps the unit circle, remembering that Möbius transforms map circles and lines to circles and lines.