Re: complex equation
- To: mathgroup at smc.vnet.net
- Subject: [mg118535] Re: complex equation
- From: Christopher Arthur <aarthur at tx.rr.com>
- Date: Mon, 2 May 2011 06:50:47 -0400 (EDT)
This is a good answer for college algebra. Often with these kinds of
equations our checking is necessary!
Gary Wardall a =E9crit :
> On Apr 27, 4:38 am, Antonio Mezzacapo <ant.mezzac... at 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
>
>
>