Math Is Fun Forum

simron

Real Member

Registered: 2006-10-07

Posts: 237

Imaginary number root

Recently, I got thinking about what

was. I then found out that

Doesn't that seem a little odd to you?
(And as a footnote, could someone check my LaTeX?)

You were missing a } on line 2 after \frac{1}{i} - Ricky

Last edited by simron (2008-06-19 05:50:44)

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Imaginary number root

Recently, I got thinking about what

was. I then found out that

Doesn't that seem a little odd to you?
(And as a footnote, could someone check my LaTeX?)

Fixed your LaTeX, you just had a } missing in one of the indices.
I don't think the third step is allowed though.

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It wanted to be normal.

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Imaginary number root

It's definitely wrong.  Consider this simple example:

According to your method this is equal to

However, it's easy to see that in fact

All that your rearranging accomplished was to show that

which we already knew.

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mthw2vc

Member

Registered: 2008-02-15

Posts: 1

Re: Imaginary number root

The ith root of i is e^(pi/2)

mthw2vc

Member

Registered: 2008-02-15

Posts: 1

Re: Imaginary number root

My problem is with the fact that x^i is ith root (1/x) is (x^-i)^(1/i)=(x^-i)^-i*=(x^-1)**=(1/x)
when this is done twice, we get something like x=x^-1
x=1/x
This is only true for x = {1,-1}

This makes NO sense whatsoever.

*1/(a+bi)=a/(a^2 + b^2) - (b/(a^2 + b^2))i
**(a^b)^c=a^(b*c)

(Got rid of those pesky smilies)

Last edited by mathsyperson (2008-11-10 23:22:55)

mthw2vc

Member

Registered: 2008-02-15

Posts: 1

Re: Imaginary number root

Thus, I choose to avoid complex powers.