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Topic: Is this a valid relation?
Replies: 6   Last Post: Jul 8, 2013 3:25 PM

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G. A. Edgar

Posts: 2,497
Registered: 12/8/04
Re: Is this a valid relation?
Posted: Jul 7, 2013 8:23 AM
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In article <b3r46iFin8vU1@mid.individual.net>, Axel Vogt
<&noreply@axelvogt.de> wrote:

> On 06.07.2013 18:42, clicliclic@freenet.de wrote:
> >
> > "G. A. Edgar" schrieb:

> >>
> >> In article <51D7F606.6FC3A277@freenet.de>, <clicliclic@freenet.de>
> >> wrote:
> >>

> >>> Let u,v be arbitrary complex numbers and #i the imaginary unit. Can
> >>> you prove or disprove the relation
> >>>
> >>> SQRT(u - #i*v)*(u + v + #i*SQRT(u^2 + v^2)) =
> >>> SQRT(u + #i*v)*(SQRT(u^2 + v^2) + v + #i*u)
> >>>
> >>> with the help of your favorite CAS?
> >>>

> >>
> >> False as stated, needs an extra #i in there...
> >>

> >
> > Indeed, for u = 1+#i and v = 1-2*#i, for example, the equation becomes:
> >
> > 1 - SQRT(2*SQRT(13) - 6)/2 + #i*(2 + SQRT(2*SQRT(13) + 6)/2) =
> > 2 + SQRT(2*SQRT(13) - 6)/2 - #i*(3 + SQRT(2*SQRT(13) + 6)/2)
> >
> > or
> >
> > 0.4497494773 + 3.817354021*#i = 2.550250522 - 4.817354021*#i
> >
> > But I do not know how to repair this ...
> >
> > Martin.
> >

>
> But: what do you expect as re-formulation of the r h s ?


The repaired version is
SQRT(u - #i*v)*(u + #i*v + #i*SQRT(u^2 + v^2)) =
SQRT(u + #i*v)*(SQRT(u^2 + v^2) + v + #i*u)

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/



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