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In the case of the square root of a negative real (say, the discriminant in a quadratic equation), there are considered both signs of the square root with 'i' (denoting the imaginary quantity) being outside the radical, i.e. $ x_{1,2}=\frac{-b\pm i\sqrt{\Delta}}{2a}$.

But for a complex quantity inside the square root, the negative sign is ignored, as it is stated that the complex quantity's square root will generate both the negative and positive quantities. So, for solving a quadratic equation with complex number as the discriminant, the roots are stated as $x_{1,2}=\frac{-b+\sqrt{\Delta}}{2a}$.

I am confused by the latter statement about the complex number, and also want a generalization for any root value, say cube root of a complex number, and so on.

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  • $\begingroup$ If you are following a syllabus recommending to use the notation $$\sqrt{z}$$ for $z$ complex, not real nonnegative, well... I can only express my sympathy, but you might try to get some more serious sources on the subject. $\endgroup$ – Did Sep 24 '17 at 18:31
  • $\begingroup$ Do you know how to derive the quadratic formula? Do it while keeping in mind that the numbers you're working on are complex, and I hope you'll find that the answer will come to you. $\endgroup$ – Arthur Sep 24 '17 at 18:33
  • $\begingroup$ @Did: Thanks for insight into the symbols used for that case. I hope that you meant that square root symbol is not meant for complex quantity. If not, please indicate more. $\endgroup$ – jiten Sep 24 '17 at 18:34
  • $\begingroup$ ?? Why do you hypothetize that I would be meaning something else than what is written in my comment? $\endgroup$ – Did Sep 24 '17 at 18:37
  • $\begingroup$ @Arthur: It is easy to derive it, given the equation $ax^2 + bx + c$ => $(x^2 + \dfrac{b}{a} + \dfrac{b^2}{4a^2}$) = $\frac {b^2}{4a^2} -\dfrac{c}{a}$ => $(x+\dfrac{b}{2a})^2$=$\dfrac{b^2-4ac}{4a^2}$. Now this leads to the quadratic form. If any of the term a, b, or c is imaginary the discriminant is imaginary. But, it doesn't go further than that in terms of realizing the answer. $\endgroup$ – jiten Sep 24 '17 at 19:02
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What happens is the $x\in[0,+\infty)$, the symbol $\sqrt x$ means the only square root of $x$ which belongs to $[0,+\infty)$. So, if we want to talk about both square roots of $x$, we must mention $\sqrt x$ and $-\sqrt x$,

However, if $x\in\mathbb{C}\setminus[0,+\infty)$, there is no standard choice among the square roots of $x$. So, the symbol $\sqrt x$ may mean any square root of $x$.

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That statement is really inaccurate. To be more precise, it is a fact that every nonzero complex number (including all nonzero reals) has exactly two square roots, and they are opposites.

They are square roots in the sense that their squares are both equal to the given nonzero complex number. We often avoid using the symbol "$\sqrt{\cdot}$" in this context since it is no longer natural which root should be assigned to that symbol. But if you do decide which root is $\sqrt{z}$, the other root is always $-\sqrt{z}$.

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