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#1 2008-06-19 05:47:42

simron
Real Member
Registered: 2006-10-07
Posts: 237

Imaginary number root

Recently, I got thinking about what


was. I then found out that

what 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)


Linux FTW

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#2 2008-06-19 06:26:09

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Imaginary number root

simron wrote:

Recently, I got thinking about what


was. I then found out that

what 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.


Why did the vector cross the road?
It wanted to be normal.

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#3 2008-06-19 06:33:39

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.


Wrap it in bacon

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#4 2008-11-10 13:39:23

mthw2vc
Member
Registered: 2008-02-15
Posts: 1

Re: Imaginary number root

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

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#5 2008-11-10 13:53:07

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)

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#6 2008-11-10 13:55:13

mthw2vc
Member
Registered: 2008-02-15
Posts: 1

Re: Imaginary number root

Thus, I choose to avoid complex powers. down

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