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Here is my derivation.

$i^2= ((-1)^{0.5})^2$

$=((-1)^2)^{0.5}$

= $1$

Is there a rule saying that 2nd line is not allowed? I have seen on a related post that sqrt(a).sqrt(b)=sqrt(a.b) is true only if a and b are above zero But i do not understand what it can mean for a and b (2 complex numbers) to be greater than zero? Please explain this rule. And also if the rule can be more generally stated?

Note: i learnt in school that square root of x means the POSITIVE square root of x. That was true throughout all algebra before we got to complex numbers. I now understand that that referred to the "principal" root. (Root with the highest real element).

Also when we define i as square root of -1, this is a strange statement actually because also, -i is also sqr rt of -1.

Randor
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  • You don't have $\sqrt{a}\sqrt{b}=\sqrt{ab}$ in general for complex numbers. –  Feb 15 '21 at 21:48
  • Going from line 2 to 3, there are two square roots of 1, the negative root is the correct solution to your equation and the positive root is called an extraneous root. – John_Krampf Feb 15 '21 at 21:50
  • Essentially, you should avoid blindly applying power rules. Also, saying things like "${x^{0.5}}$" is ambiguous - do you mean the principal square root? Or are you treating ${x^{0.5}}$ as the set of all values $y$ such that ${y^2=x}$? If you are doing the latter - as @John_Krampf said - notice ${(-1)^2=1}$, and so ${-1 \in 1^{0.5}}$ – Riemann'sPointyNose Feb 15 '21 at 21:52
  • Its a pity this question was closed. If my question is ambiguous then please restate my question using mathematics that are not ambiguous – Randor Feb 15 '21 at 22:13
  • Cf. this question – J. W. Tanner Feb 15 '21 at 22:17

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