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Surprising answer III (Posted on 2023-06-04) Difficulty: 2 of 5
Evaluate:

sqrt(i)+sqrt(-i).

o There are at least 3 distinct methods

No Solution Yet Submitted by Ady TZIDON    
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Some Thoughts re: 2nd method - principal roots argument | Comment 3 of 6 |
(In reply to 2nd method by Larry)

I had a similar train of thought.  But then I stepped back a bit. Let's consider simplifying sqrt(9)+sqrt(4).  This is basically trivial, as it reduces to 3+2=5.  


But the only reason this expression is trivial is because we have a well-defined commonly accepted definition of the Principal Square Root over positive reals - always choose the positive root of the two possible roots.

For complex numbers, one convention is that of the roots of nthroot(z) then the Principal nthroot(z) is the one with the smallest angle measured counterclockwise from the positive real axis.  When restricted to square roots then this convention effectively says to choose the square root with the positive complex part.  In polar form the principal square root will have 0<=theta<pi.

Then by this convention the principal root of sqrt(i) is 1/sqrt(2) + i/sqrt(2) and the principal root of sqrt(-i) is -1/sqrt(2) + i/sqrt(2).
Then sqrt(i) + sqrt(-i) = (1/sqrt(2) + i/sqrt(2)) + (-1/sqrt(2) + i/sqrt(2)) = i*sqrt(2).

Another very similar rule is to take the principal square root as the one with the smallest angle to the real axis in either direction, which is the same as choosing the root with the positive real part. In polar form the principal square root will have -pi/2<theta<=pi/2.

In this case we'll get sqrt(i) + sqrt(-i) = (1/sqrt(2) + i/sqrt(2)) + (1/sqrt(2) - i/sqrt(2)) = sqrt(2).

Edited on June 4, 2023, 12:13 pm
  Posted by Brian Smith on 2023-06-04 12:12:24

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