Solving sqrt(i)=z, one or two solutions?

[Nigel (UK)]

(10-25-2018 07:17 AM)sasa Wrote:  Another interesting example...

Solve: \( \sqrt{i} = z \)

Not at all complicated task, it have two solutions. However, we will get interesting results from the Prime, depending how we formulate it for solving.

Let try first simply \( \sqrt{i} \), it returns \( \frac{1+i}{\sqrt{2}} \)
Let try further solve((sqrt(i)) = z,z), result is \( \left \{ \frac{1+i}{\sqrt{2}} \right \} \)
However: solve((i) = (z^2),z), result is \( \left \{ \frac{1+i}{\sqrt{2}}, -{\frac{1+i}{\sqrt{2}}} \right \} \)

An obvious path for lower grade students to solve it, is to use elementary definition of a complex number. Another solution can be using Euler's formula...

However, the Prime unexpectedly gave different number of solutions, depending on how formula is written...

Disclaimer: Tested on latest public beta emulator only.

I expect that the Prime treats the square root function as a single-valued function, so that (for example) \(\sqrt4=+2\) only and solving \(\sqrt4=x\) only returns \(x=2\) as a solution. The Prime views the equations \(\sqrt4=x\) and \(4=x^2\) as two different equations with one and two roots respectively; I think that this is the behaviour I would expect. What do others feel?

Nigel (UK)