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Re: complex equation

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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...@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
>
>
>  

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Re: complex equation Christopher Arthur <aarthur@tx.rr.com> - 2011-05-02 10:50 +0000

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