Re: Why is this evaluation is not giving me the same solution

Roger Stafford wrote:
>
>
> In article <ef2009e.-1@xxxxxxxxxxxxxxxx>, Koang
> <kthich@xxxxxxxxxxx> wrote:
>
>> I have been struggling with the evaluation of imaginary numbers
> and I
>> think what am experiencing is getting me nowhere. Here is the
>> problem. I am trying to find the quotient of imaginary numbers.
>>
>> a=i*sqrt(i);
>> b=sqrt(-i);
>> c=a/a = 1;
>> d=a/b = -1;
>>
>> I thought that sqrt(-i)=sqrt(i^3)=i*sqrt(i);
>>
>> My question is why am I not getting the same answer? I thought
I
>> should get the same answer no matter what since
>> sqrt(-i)=sqrt(i^3)=i*sqrt(i). I try all math software (mathlab,
>> mathcad, maple) and not surprisingly enough the answer to c and
d
> are
>> different. Is there any explaination?
>>
>> I would highly welcome any feedback
>>
>> Koang
> ----------------------
> Hello Koang,
>
> Your mistake comes at the step, sqrt(i^3) = i*sqrt(i), in trying
> to
> remove the i^2 factor from inside the square root. There is a
> similar
> anomaly with the following reasoning:
>
> 2 = sqrt(4) = sqrt((-1)^2*4) = (-1)*sqrt(4) = -2
>
> For numbers that are real and positive, it is easy to decide that
> all
> square roots will be chosen positive, and to avoid taking the
> square root
> of negative numbers. In the complex plane it is somewhat more
> difficult
> to decide which of two possible values to give as square root. If
> you
> commence at the real value z = 1, starting with a sqrt(z) value of
> 1, and
> then travel around the origin counterclockwise in a circle through
> 360
> degrees, always making sure that the square root remains a
> continuous
> function, you will be forced to arrive back at z = 1 with a square
> root
> value of -1! If you travel around the circle once again, you will
> return
> to a square root value of +1. This gives you a hint of what is
> meant by
> the statement that the analytic function, square root, has two
> "branches". This gives rise to all sorts of anomalies such as the
> one you
> have produced.
>
> Matlab gives only a "principal" value for the 'sqrt' function,
> which
> always chooses the square root that has the a non-negative value
> for its
> real part. Hence,
>
> sqrt(i^3) = sqrt(-i) = 0.7071 - 0.7071i
> i * sqrt(i) = i * (0.7071 + 0.7071i) = -0.7071 + 0.7071i
>
> There is no way out of this dilemma, once you have chosen that
> convention
> for a principal value. Furthermore any other convention would be
> bound to
> run into similar paradoxes. It is important to recognize the
> presence of
> two values on the two branches for every square root even if only
> one of
> them is produced as a principal value.
>
> (Remove "xyzzy" and ".invalid" to send me email.)
> Roger Stafford
>

Roger Stafford,

That make alot of sense. So, I shouldn't simplify sqrt(-i) in order
to get a correct answer right? Thank you.

Koang
.

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