Evaluate:
sqrt(i)+sqrt(-i).
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There are at least 3 distinct methods
(In reply to
re(2): 2nd method - principal roots argument ? by Larry)
In a sense, yes. I'm not an expert on this stuff, but I'll explain what I can.
There is a class of relations referred to as multivalued functions, where the relation outputs a well-defined set of possible values for its inputs. The square root is the simplest of the multivalued functions.
So if we use the broad multivalued definition of sqrt(z), then each root puts out two values in their respective codomains, and then we need to take all the possible permutations to generate all the possible solutions. That is pretty much what you did in your solution.
So then what gets us from a multivalued function into a proper function? Usually that is done by applying a branch cut and calling a chosen branch the principal value.
For sqrt(x) in reals, its pretty easy to look at the relation y^2=x and see two branches are the positive half and the negative half of the parabola. So it is straightforward to take the positive branch of sqrt(x) to be the principal branch because then the domain and codomain are both the set of nonnegative reals.
For sqrt(z) over complex numbers, its not as straightforward as there are many possible ways make a branch cut. The two I presented are the ones I have seen, and I believe the second one is the closest thing to an "official" principal value of sqrt(z) for complex numbers.
In summary, if we use the multivalued definition of sqrt(z) then we get four valid answers to the original question. But as soon as we introduce a branch cut to define the principal sqrt as a proper function then three of the answers are discarded leaving us with just one answer; and which answer remains depends on which branch cut we take.
re(3): 2nd method - principal roots argument ?
