Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]
Groups > comp.soft-sys.math.mathematica > #1964
| From | Gary Wardall <gwardall@gmail.com> |
|---|---|
| Newsgroups | comp.soft-sys.math.mathematica |
| Subject | Re: complex equation |
| Date | 2011-04-28 10:36 +0000 |
| Organization | Steven M. Christensen and Associates, Inc and MathTensor, Inc. |
| Message-ID | <ipbfv7$afn$1@smc.vnet.net> (permalink) |
| References | <ip8o5p$pfi$1@smc.vnet.net> |
On Apr 27, 4:38 am, Antonio Mezzacapo <ant.mezzac...@gmail.com> wrote:
> Hi everyone,
> I have a question. Do you know why Mathematica finds solution to this equation
> Solve[Sqrt[1 - z^2] = 2, z]
> {{z -> -I Sqrt[3]}, {z -> I Sqrt[3]}}
> while if I change sign of the right part it doesn't find solution anymore?
> Solve[Sqrt[1 - z^2] = -2, z]
> {}
>
> Is this related to the phase specification for complex numbers?
>
> Thank you
> Antonio Mezzacapo
Antonio,
Sqrt[1 - z^2] = -2
has no solutions, real or complex. That is the solution set is empty.
Mathematica is correct when it yields {}.
Note:
Sqrt[1 - z^2] = -2
(Sqrt[1 - z^2] )^2 = (-2)^2
1 - z^2 = 4
- z^2 = 3
z^2 =- 3
z = -i Sqrt[3] or z= i Sqrt[3]
Checking/Proving:
Sqrt[1 - (-i Sqrt[3])^2] = -2
2 = -2 NO!
Sqrt[1 - (i Sqrt[3])^2] == -2
2 = -2 NO!
Gary Wardall
Back to comp.soft-sys.math.mathematica | Previous | Next — Previous in thread | Find similar | Unroll thread
complex equation Antonio Mezzacapo <ant.mezzacapo@gmail.com> - 2011-04-27 09:38 +0000 Re: complex equation Gary Wardall <gwardall@gmail.com> - 2011-04-28 10:36 +0000
csiph-web