5
$\begingroup$

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

Don't know where to start.. Please help.!

NOTE: $z$ is a complex number.

$\endgroup$
1

3 Answers 3

11
$\begingroup$

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 $\cos2A=1-2\sin^2A$)

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

$\endgroup$
3
  • $\begingroup$ Now what can we conclude from this ? $\endgroup$
    – Apurv
    Dec 15, 2013 at 7:02
  • $\begingroup$ @Apurv, so if $z=x+iy,x=-\frac12.$ What can we conclude from here? $\endgroup$ Dec 15, 2013 at 7:03
  • $\begingroup$ Ok.. So the points lie on x=-1/2. Hence the points are collinear. Thanks !! $\endgroup$
    – Apurv
    Dec 15, 2013 at 7:06
7
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.