Hi, my question is :

Let n be a positive integer. Find all real numbers x such that $\displaystyle (x-i)^n + (x+i)^n = 0 $

I find that there is no such REAL number.

Does someone find a different answer?

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- Jan 13th 2009, 04:52 PMvincisonfireComplex Number
Hi, my question is :

Let n be a positive integer. Find all real numbers x such that $\displaystyle (x-i)^n + (x+i)^n = 0 $

I find that there is no such REAL number.

Does someone find a different answer? - Jan 13th 2009, 05:03 PMProve It
- Jan 13th 2009, 05:09 PMMush
Well a pretty obvious solution from inspection is that if n is even, then x = 0 is a real number solution. But I'll do the working just to see if any others lurk:

$\displaystyle (x-i)^n = -(x+i)^n $

$\displaystyle x-i = (-(x+i)^n)^{\frac{1}{n}} $

$\displaystyle x-i = (-1)^{\frac{1}{n}} (x+i) $

$\displaystyle x-i = \frac{(-1)^1}{(-1)^n} \times (x+i) $

$\displaystyle x-i = \frac{-1}{(-1)^n} \times (x+i) $

$\displaystyle x-i = \frac{-x}{(-1)^n} - \frac{i}{(-1)^n} $

Get all x terms on LHS and non-x terms on RHS

$\displaystyle x+ \frac{x}{(-1)^n} = i- \frac{i}{(-1)^n} $

$\displaystyle x(1+ \frac{1}{(-1)^n}) = \frac{i(-1)^n-i}{(-1)^n} $

$\displaystyle x(\frac{(-1)^n+1}{(-1)^n}) = \frac{i(-1)^n-i}{(-1)^n} $

$\displaystyle x = \frac{i(-1)^n-i}{(-1)^n} \times \frac{(-1)^n}{(-1)^n+1} $

$\displaystyle x = \frac{i(-1)^n-i}{(-1)^n+1} $

$\displaystyle x = \frac{i((-1)^n-1)}{(-1)^n+1} $

Hence $\displaystyle n \neq 2k+1 $ for $\displaystyle k$ any positive integer.

For $\displaystyle n = 2k$ $\displaystyle x = 0 $

Hence the equation has a real number solution for all even values of n. And that solution is $\displaystyle x = 0$. 0 is a real number. - Jan 13th 2009, 05:10 PMvincisonfire
Shouldn't it be $\displaystyle

2(x^2 + 1)^{\frac{n}{2}}\cos{(\frac{n}{x})} = 0

$ because the angle will be $\displaystyle arctan\frac{1}{x} $ - Jan 13th 2009, 05:11 PMProve It
- Jan 13th 2009, 05:19 PMvincisonfire
I would thus have $\displaystyle x = \frac{2n}{(2k+1)\pi}, k\in \mathbb Z $ but it does not work.

And thanks Mush. I guess there is no solution for all n. - Jan 13th 2009, 05:37 PMNonCommAlg
the solutions are $\displaystyle x=\cot \left(\frac{(2k+1)\pi}{2n} \right), \ 1 \leq k \leq n.$ just put $\displaystyle x=\cot \alpha$ and solve a simple equation.

- Jan 13th 2009, 05:45 PMvincisonfire
Mush somewhere you did $\displaystyle *(-1)^{\frac{1}{n}}=\frac{-1}{(-1)^n} $*

- Jan 13th 2009, 05:53 PMProve It
- Jan 13th 2009, 05:56 PMvincisonfire
To NonCommAlg :

I don't understand why but as n goes up the solutions are not working???

e.g. when n = 101 mathematica gives me 2.59828*10^167 + 0. I - Jan 13th 2009, 06:00 PMvincisonfire
Yes it should be $\displaystyle

\sqrt[n]{-1}

$. That complicates the things ... but the answer is still good. - Jan 13th 2009, 06:05 PMNonCommAlg
- Jan 13th 2009, 06:09 PMvincisonfire
No I meant $\displaystyle (Cot(\frac{\pi}{202})+i)^{101} + (Cot(\frac{\pi}{202})-i)^{101} = 2.59828*10^{167} + 0. I $

When n > 10 it begins to have great values. - Jan 13th 2009, 06:14 PMProve It
OK this is driving me mad.

According to my CAS, it reduces to...

$\displaystyle 2\cos{\left(n\arctan{x} - \frac{n\pi}{2}\right)}(x^2 + 1)^{\frac{n}{2}}$

and gives

$\displaystyle x = -\cot{\left(\frac{k\pi}{n} - \frac{\pi}{2n}\right)}$ and $\displaystyle -1 \leq \frac{n + 2k - 1}{n} \leq 1$. - Jan 13th 2009, 06:16 PMNonCommAlg