?>
I want to do something for this site's 10th birthday in a month and ten days

# Department of Science, Math, & Technology

E = mcï¿½
A simple complex number equation
Posted: Posted July 18th
Edited July 18th by The Fly
 As usual, i denotes the complex number square root of -1. (So (i^2 = -1).) For m,n > 0 odd coprime integers, can you express the fraction $\large \frac{i^m - 1}{i^n - 1}$ as a power of i? It turns out that it is a power of i, and we are to find what that power is as a function of m,n. Good luck.
There are 6 Replies

It appears that the Latex commands have been lost during the transition? So my latex math isn't showing up as it use to. Don't know if you can read my equations above.

It looks like the blackslash command doesn't work anymore - at least when I try to edit. Aha, you have to double the backslash like \\

Testing:

$\left( \sum_{k=1}^n a_k b_k \right)^{!!2} \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$

Edited July 18th by The Fly
Edit Filter Quote Report

The backslash command \ for Latex disappears when you re-edit a post to correct it.

Posted July 18th by The Fly
Edit Filter Quote Report

I found a general solution, but it isn't a single formula -- I can't figure out how to piece the data together into one.

I started with the idea that, well, -1 is just i2:

$\large \frac{i^m + i^2}{i^n + i^2}$

I then did a lot of unnecessary work, and eventually realized that you can just modulus m or n by four because i5 is just i*i4, or put another way, it's 1*i, which is just i.

This means that m and n collapse to either 1 or 3.

I noticed that you can reduce i1 and i3 in this way:

$\large \frac{i^3 + i^2}{i^1 + i^2}$

$\large \frac{i^2 ( i + 1 )}{i ( i + 1 ) }$

The i+i cancels out, and you're left with some power of i.

I then looked all cases, changing 1 to i4 where appropriate:

• If m is 1 and n is 3, the answer is i-1
• If m is 3 and n is 1, the answer is i1
• If m is 1 and n is 1, the answer is i4
• If m is 3 and n is 3, the answer is i4

Given more time, I could probably reduce this into a single formula (since that's kinda been my specialty lately), but I couldn't quite figure out how to put it together.

• Edited July 18th by Xhin
Edit Filter Quote Report
Xhin
Sky's the limit

That backslash thing is extremely annoying. Will fix.

Posted July 18th by Xhin
Edit Filter Quote Report
Xhin
Sky's the limit

Nevermind, figured it out.

i[ (m%4)-(n%4) ]/2

Basically, this takes advantage of the fact that i0 is the same as i4

Edited July 18th by Xhin
Edit Filter Quote Report
Xhin
Sky's the limit

If n is 0 or a multiple of 4 the denominator is 0 so the fraction ((i^m)-1)/0 is meaningless; not a power of i.
Otherwise, if m is 0 or a multiple of 4 the numerator is 0 so the fraction 0/((i^n)-1) is 0. This is not a power of i.
Otherwise if m and n are congruent modulo 4 then the numerator and denominator are equal so the fraction has value 1. This is a power of i.
Otherwise you have either (i-1)/(-i-1) or (-i-1)/(i-1)
-i = i^3 so (-i-1)/(i-1)=((i^3)-1)/(i-1)=(i^2)+i+1 = -1 + i +1 = i.
(i-1)/(-i-1) is 1/i is -i. These last two values are both powers of i.
The fraction never evaluates to -1. That’s the only power of i to which the expression does not evaluate for any m or n.

Can you calculate i^i ?
Can you prove it’s a real number?

Edited July 25th by chiarizio
Edit Filter Quote Report
Reply to: A simple complex number equation