# Complex Number

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• January 13th 2009, 04:52 PM
vincisonfire
Complex Number
Hi, my question is :
Let n be a positive integer. Find all real numbers x such that $(x-i)^n + (x+i)^n = 0$
I find that there is no such REAL number.
Does someone find a different answer?
• January 13th 2009, 05:03 PM
Prove It
Quote:

Originally Posted by vincisonfire
Hi, my question is :
Let n be a positive integer. Find all real numbers x such that $(x-i)^n + (x+i)^n = 0$
I find that there is no such REAL number.
Does someone find a different answer?

Hint: convert to polars and use De Moivre's Theorem.

You should find that it reduces down to

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

You should be able to solve this with null factor law.
• January 13th 2009, 05:09 PM
Mush
Quote:

Originally Posted by vincisonfire
Hi, my question is :
Let n be a positive integer. Find all real numbers x such that $(x-i)^n + (x+i)^n = 0$
I find that there is no such REAL number.
Does someone find a different answer?

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:
$(x-i)^n = -(x+i)^n$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Hence the equation has a real number solution for all even values of n. And that solution is $x = 0$. 0 is a real number.
• January 13th 2009, 05:10 PM
vincisonfire
Shouldn't it be $

2(x^2 + 1)^{\frac{n}{2}}\cos{(\frac{n}{x})} = 0
$
because the angle will be $arctan\frac{1}{x}$
• January 13th 2009, 05:11 PM
Prove It
Quote:

Originally Posted by vincisonfire
Shouldn't it be $

2(x^2 + 1)^{\frac{n}{2}}\cos{(\frac{n}{x})} = 0
$
because the angle will be $*arctan\frac{1}{x}$

Yes you're right, my apologies.

I did $\frac{\Re{z}}{\Im{z}}$ instead of $\frac{\Im{z}}{\Re{z}}$.
• January 13th 2009, 05:19 PM
vincisonfire
I would thus have $x = \frac{2n}{(2k+1)\pi}, k\in \mathbb Z$ but it does not work.
And thanks Mush. I guess there is no solution for all n.
• January 13th 2009, 05:37 PM
NonCommAlg
the solutions are $x=\cot \left(\frac{(2k+1)\pi}{2n} \right), \ 1 \leq k \leq n.$ just put $x=\cot \alpha$ and solve a simple equation.
• January 13th 2009, 05:45 PM
vincisonfire
Mush somewhere you did $*(-1)^{\frac{1}{n}}=\frac{-1}{(-1)^n}$*
• January 13th 2009, 05:53 PM
Prove It
Quote:

Originally Posted by vincisonfire
Mush somewhere you did $*(-1)^{\frac{1}{n}}=\frac{-1}{(-1)^n}$*

I was wondering about that too.

Shouldn't it be $\sqrt[n]{-1}$?
• January 13th 2009, 05:56 PM
vincisonfire
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
• January 13th 2009, 06:00 PM
vincisonfire
Yes it should be $

\sqrt[n]{-1}
$
. That complicates the things ... but the answer is still good.
• January 13th 2009, 06:05 PM
NonCommAlg
Quote:

Originally Posted by vincisonfire
To NonCommAlg :
I don't understand why but as n goes up the solutions are not working???

the solution still work but you won't get any new ones because: $\cot \left(\frac{(2k+1)\pi}{2n} \right)=\cot \left(\frac{(2(k+n)+1)\pi}{2n} \right).$
• January 13th 2009, 06:09 PM
vincisonfire
No I meant $(Cot(\frac{\pi}{202})+i)^{101} + (Cot(\frac{\pi}{202})-i)^{101} = 2.59828*10^{167} + 0. I$
When n > 10 it begins to have great values.
• January 13th 2009, 06:14 PM
Prove It
OK this is driving me mad.

According to my CAS, it reduces to...

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

and gives

$x = -\cot{\left(\frac{k\pi}{n} - \frac{\pi}{2n}\right)}$ and $-1 \leq \frac{n + 2k - 1}{n} \leq 1$.
• January 13th 2009, 06:16 PM
NonCommAlg
Quote:

Originally Posted by vincisonfire
No I meant $(Cot(\frac{\pi}{202})+i)^{101} + (Cot(\frac{\pi}{202})-i)^{101} = 2.59828*10^{167} + 0. I$
When n > 10 it begins to have great values.

the solution i gave you is exact. you trust the software you're using more than mathematics?!!
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