# Prove that $f=(x+i)^{10}+(x-i)^{10}$ have all real roots

We have $f=(x+i)^{10}+(x-i)^{10}$ and we need to prove that $f$ have all the roots in $\mathbb{R}$.

Here is all my steps:

• Suppose that $z\in\mathbb{R}$ is a root of $f\Rightarrow (z+i)^{10}+(z-i)^{10}=0$

Therefore: $f(z)=\sum_{k=0}^{10}\left[\left(\dbinom{10}{k}\cdot z^{10-k}\cdot i^k\right)\left(1+(-1)^k\right)\right]=0$

I don't have ideea how can I prove that $f$ have all real roots

Suppose that $z=a+bi$ and we need to prove that $b=0$:

$$\Rightarrow f(z)=\sum_{k=0}^{10}\left(\dbinom{10}{k}\cdot a^{10-k}\cdot i^k\right) \left[(b+1)^k+(b-1)^k \right]=0$$

$$\Rightarrow \left[(b+1)^k+(b-1)^k \right]=0$$

$$\Rightarrow b+1=-b+1$$ $$\Rightarrow b=0$$

Therefore $f$ have all the roots in $\mathbb{R}$

• Here is a photo with the proof of real axis, I don't know if is correct: $(x+i)^{10}+(x-i)^{10}=0$ immediately implies $|x+i|=|x-i|$, which is equivalent to $x \in \mathbb R$ (the real axis is the perpendicular bisector of the line from $i$ to $-i$).

• can you give me a draw ? because I don't know to make it Jun 5, 2015 at 8:04
• What exactly you do not know/understand about this proof? Be more precise.
– MooS
Jun 5, 2015 at 8:07
• MooS I can't understand your method, can you give some explanations ? Jun 5, 2015 at 8:11
• @Lucas: |x+i| is the distance of $x$ from $-i$ and $|x-i|$ is the distance of $x$ from $i$. The locus of points that are equidistant from $-i$ and $i$ is the real axis.
– robjohn
Jun 5, 2015 at 11:44
• All roots satisfy $|x-i|=|x+i|$. So this works fine without further arguments.
– MooS
Jun 5, 2015 at 18:41

$(x+i)^{10} = -(x-i)^{10} = i^{10}(x-i)^{10} = (i(x-i))^{10} \to \left(\dfrac{x+i}{ix+1}\right)^{10} = 1$. Can you continue??

• notaloner I can't understand how you obtain $\left(\frac{x+i}{1+ix}\right)^{10}$ and I can't continue Jun 5, 2015 at 8:23
• first, $i(x-i) = ix - i^2 = ix+1$, and secondly use $a^{10} = b^{10} \iff \left(\dfrac{a}{b}\right)^{10} = 1$ Jun 5, 2015 at 8:25
• NotALoner if I continue your method I obtain something like this $\frac{(x+i)(ix-1)}{-x^2-1}=1\Rightarrow x^2(-1-i)+2x+i-1=0$ and I suppose we need to put condition that: $4-4(-1-i)(i-1)\geq 0$ isn't it ? Jun 5, 2015 at 8:47
• after I obtain: $-4\geq 0$ which is a contradiction Jun 5, 2015 at 8:51
• can you explain what you want by this method ? Jun 5, 2015 at 8:57

Let $a+ib$ be an imaginary root of $f$ , where , $a,b\in \mathbb R$. Then $$f(a+ib)=0$$

$$\implies (a+i(b+1))^{10}+(a+i(b-1))^{10}=0$$

Expand two binomial series , and separate real part and imaginary parts. Put them separately $0$. Then you obtain the value of $b$, which would be zero, if the problem is proper.

• S.Panja from here I have to expand $(a+i(b+1))^{10}$ in binominal series? Jun 5, 2015 at 8:02
• I cancel your downvote Jun 5, 2015 at 8:08
• Panja I will edit with binominal series and I want to tell me if is correct Jun 5, 2015 at 8:09
• Panja is correct ? Jun 5, 2015 at 8:32

Here is another way of showing that all solution are real valued.

$(x+i)^{10}+(x-i)^{10}=0\implies\,\,\left(\frac{x+i}{x-i}\right)^{10}=e^{in\pi}\implies \frac{x+i}{x-i}=e^{in\pi/10} \implies x=\cot(n\pi/20)$

which are real-valued for all integer $n$.

To answer the question regarding the substitution $z=a+ib$ and subsequent analysis, if $(b+1)^k+(b-1)^k=0$, then this implies that $b=-i\cot (n\pi/2k)$ for all integer $n$ and $k=0, 1, 2, \cdots 10$.

But, since $b$ is presumed to be real valued, then this implies $b=0$ (i.e., only solutions with $n/k$ odd are valid).

However, this assumption (i.e., $(b+1)^k+(b-1)^k=0$) is not the basis for a proof that $b=0$.

• @Lucas I'm not sure that this is what you were looking for, but hope that it helps. Jun 5, 2015 at 17:35
• Dr.Mv why $(x+i)^{10}+(x-i)^{10}$ leds to absolute value: $|x+i|=|x-i|$ ? Jun 5, 2015 at 17:44
• @Lucas Well, that result was not used here. But you recall that $|z|^n=|z^n|$. So, letting $z=\frac{x+i}{x-i}$, we have $|z^{10}|=|-1| \implies |z|^{10}=1\implies |z|=1$ is one root. But there are other roots. Jun 5, 2015 at 17:49
• Dr.Mv how can I continue my solution with binominal series ? Jun 5, 2015 at 17:52

You can transform the equation in

$$\left(\frac{x+i}{x-i}\right)^{10}=-1=z^{10}.$$

Then, solving for $x$ and using $|z|=1$, $$x=i\frac{z+1}{z-1}=i\frac{(z+1)(\overline z-1)}{(z-1)(\overline z-1)}=i\frac{|z|^2-1+\overline z-z}{|z|^2+1-z-\overline z}=\frac{2\Im z}{2-2\Re z}.$$

• my binominal series is correct ? Jun 5, 2015 at 9:32
• Yes but the conclusion you draw from it is not.
– user65203
Jun 5, 2015 at 9:35
• and how can I continue at final ? Jun 5, 2015 at 9:40
• Leads you to a system of two polynomial equations in two unknowns.
– user65203
Jun 5, 2015 at 9:48
• and which is two polynomial equations ? Jun 5, 2015 at 9:50

$x=a+bi$

$$(x+i)^{10}+(x-i)^{10}=0 \Rightarrow (x+i)^{10}=-(x-i)^{10} \Rightarrow (\frac{x+i}{x-i})^{10}=-1 \\ \Rightarrow \left (\frac{a+(b+1)i}{a+(b-1)i}\right )^{10}=i^{10} \Rightarrow \frac{a+(b+1)i}{a+(b-1)i}=i \\ \Rightarrow \frac{(a+(b+1)i)(a-(b-1)i)}{(a+(b-1)i)(a-(b-1)i)}=i \Rightarrow \frac{a^2-a(b-1)i+a(b+1)i+(b^2-1)}{a^2+(b-1)^2}=i \Rightarrow \frac{(a^2+(b^2-1))+2ai}{a^2+(b-1)^2}=i$$

That means that $$\frac{a^2+(b^2-1)}{a^2+(b-1)^2}=0 \text{ and } \frac{2a}{a^2+(b-1)^2}=1$$

$$\frac{a^2+(b^2-1)}{a^2+(b-1)^2}=0 \Rightarrow a^2+b^2-1=0 \Rightarrow b^2=1-a^2$$

$$\frac{2a}{a^2+(b-1)^2}=1 \Rightarrow 2a=a^2+(b-1)^2 \Rightarrow 2a=a^2+b^2-2b+1$$

Solving for $a$ and $b$ we get the following solutions:

$a=0, b=1$ and $a=1, b=0$

We reject $(a, b)=(0, 1)$ because then the denominator would be $0$.

So, the solution is $(a, b)=(1, 0)$.

That means that $x=1 \in \mathbb{R}$.



Do the same for the case $\left (\frac{a+(b+1)i}{a+(b-1)i}\right )^{10}=i^{10} \Rightarrow \frac{a+(b+1)i}{a+(b-1)i}=-i$

• Mary Star what's wrong in my binominal series ? I can't see the real part in my solution Jun 5, 2015 at 11:47
• You can't go from $x^{10}=y^{10}$ to $x=y$ as you do in the second line of your proof here. Jun 5, 2015 at 18:42
• why it can't do it ? Jun 6, 2015 at 6:05