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I know that the equation $$ z^2=-8i $$ has two solutions, namely $z=\pm(2-2i)$. However, $\sqrt{-8i}$ is defined to be $2-2i$. Similar to the real case, $\sqrt{}$ selects the solution such that $\text{Re}(z)\geq 0$.

Why is this a convenient definition for the complex case? Is it merely the natural extension of the definition for the real case? Or is there a deeper reason? What about, for example, $$ z^4=-8i, $$ which has two solutions with positive real parts. What is the criteria here?

sam wolfe
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  • I didn't know that $\sqrt[n]z$ had an standard meaning for $z\in\Bbb C$. But perhaps the root with a lesser argument in $[0,2\pi)$ is selected. – ajotatxe Oct 02 '19 at 16:33
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    I personally avoid $\sqrt{\phantom{x}}$ (and its cousins) for complex numbers. I don't miss them, really. All they ever did was create trouble. – Arthur Oct 02 '19 at 16:56
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    What you actually mean is "someone I've talked to or read defined $\sqrt{-8i} = 2 - 2i$". As ajotatxe indicates, there is no widely-agreed-to convention for $\sqrt[n]{\phantom x}$, or more generally for $z^w$ when $w$ is not an integer and $z$ is not a positive real. Mathematicians are free to choose the definition that works best for whatever problem they are working on. One should never use such functions without explicitly saying how you are defining them. – Paul Sinclair Oct 02 '19 at 23:01

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