# Show that the real part of the root of an equation is constant

I've been stuck for a while on the following question:

Let $z$ be a root of the following equation:

$$z^n + (z+1)^n = 0$$

where $n$ is any positive integer. Show that

$$Re(z) = -\frac12$$

Because $z^n = -(z+1)^n$, I tried to write $z+1$ in terms of $z$. In Cartesian coordinates, I've tried (with $z = a + ib$):

$$(a + ib)^n = - (a + 1 + ib)^n$$

I found no way to calculate $a$ for an arbitrary $n$ in this equation. In polar coordinates:

$$\sqrt{a^2 + b^2} e^{i \, n \, atan( \frac{b}{a} )} = \sqrt{(a+1)^2 + b^2} e^{i \, n \, atan( \frac{b}{a+1} )}$$

The real part of $z$ seems hard to extract from this equation.

Any clues welcome, I've been trying is one for many hours now!

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Try to show that there are $n$ solutions of the form $$z_0 = y_0 i - \frac{1}{2}$$ Then by the fundamental theorem of algebra, you're done. – AlexR Sep 23 '13 at 12:02
@AlexR: De Moivre's theorem can prove N roots for unity, but how would that work for the equation $z^n + (z+1)^n = 0$? – Andomar Sep 23 '13 at 14:09

Since $z=0$ is not a root, our roots all satisfy $x^n =-1$ where $x=1+1/z.$ So the solutions are $$z= \frac{1}{ \cos \left(\dfrac{(1+2k)\pi}{n}\right)-1+ i\sin \left(\dfrac{(1+2k)\pi}{n}\right)}.$$

Multiply the numerator and denominator by $\cos \left(\dfrac{(1+2k)\pi}{n}\right)-1- i\sin \left(\dfrac{(1+2k)\pi}{n}\right)$ and use some basic trig identities to see that the real part is $-1/2.$

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Thanks! Would you have a hint on what thought process leads to substituting $x$ for $1 + 1/z$ ? – Andomar Sep 23 '13 at 13:16
@Andomar I wanted to get a single term with exponent $n$ so I divided both sides of the equation by $z^n.$ – Ragib Zaman Sep 23 '13 at 14:00

$$z^{n+1}=-(z+1)^{n+1}\iff z\cdot z^n=(z+1)(-(z+1)^n)\stackrel{\text{clearly}\;z\neq0}\implies \frac z{z+1}=-\left(\frac{z+1}z\right)^n$$

Putting

$$w:=\frac z{z+1}\;,\;\;\text{we got}\;\;w=-w^{-n}\iff w^{n+1}=-1=e^{\pi i}$$

The solutions of this equation are

$$w_k=e^{\pi i(2k+1)/(n+1)}\;,\;\;k=0,1,2,\ldots,n$$

and for each we have

$$w_k=\cos\frac{2\pi (2k+1)}{n+1}+i\sin\frac{2\pi(2k+1)}{n+1}$$

and $\;\color{blue}{\text{the real part of}\;\;z\;}$ is given by

$$\text{Re}\,\frac w{1-w}$$

But

$$\frac w{1-w}=\frac{w-|w|^2}{|1-w|^2}$$

and

\begin{align*}w-|w|^2&=w-1=\color{red}{\cos t-1}+i\sin t\\ |1-w|^2&=1-2\cos t+\cos^2t+\sin^2t=\color{red}{2(1-\cos t)}\;,\;\;\text{with}\;\;t=\frac{\pi(2k+1)}{n+1}\end{align*}

So, finally, we get Re$\,z=-\frac12\;$ ...

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Thanks! In the equation after "But", why is $(1 - w) (1 - \bar{w}) = |1-w|^2$ ? – Andomar Sep 23 '13 at 13:23
@Andomar, in general $\;z\in\Bbb C\implies z\overline z=|z|^2\;$ , and here $\;\overline{1-w}=1-\overline w\;$ ... – DonAntonio Sep 23 '13 at 13:27
Awesome, very instructive answer! – Andomar Sep 23 '13 at 13:49

$\underline{\bf{My Try}}$::Given $z^n+(z+1)^n = 0\Rightarrow z^n = -(z+1)^n$

Now Taking Modulus on both side, we get $\left|z^n\right| = \left|-(z+1)^n\right| = \left|(z+1)^n\right|$

$\Rightarrow \left|z\right|^n = \left|z+1\right|^n \Rightarrow \left|z-(0+i\cdot 0)\right|^n = \left|z-(-1+0\cdot i)\right|^n$

means all $P(z)=P(x,y)$ lies on the perpandicular Bisector of line joining $A(0,0)$ and $B(-1,0)$

So we can say that $\displaystyle \bf{Re(z) = -\frac{1}{2}}$

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What's a "perpandicular Bisector" ? – Andomar Sep 23 '13 at 13:17