# Branch cuts and $i$

Could anybody explain to me why after branch cuts some times $\sqrt {-1}$ is taken to be $i$ whereas at others, it is taken to be $-i$? Reference notes would also be appreciated. Thank you.

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Look at the square root function. Start at $z=-1$, and choose the value there to be $i$, then move one revolution around the origin.

If you assume continuity all the way around you would have $\sqrt{-1} = -i$, since the whole revolution in the argument variable corresponds to half a revolution of the function value. So in order for the square root to be a function (i.e. not give two different values for the same argument), it needs to be discontinious somewhere, and that somewhere is the branch cut, usually taken as the negative x-axis.

As to which value to choose, it's all the same. You try to chose a branch cut in each case so it doesn't interfere too much with whatever you are doing, and you choose one principal value. $-i$ is generally just as good as $i$, but in some cases one of them might yield prettier expressions.

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Thank you for the answer, Arthur. –  Lev Oct 10 '12 at 18:46

It all depends on what you're doing, but note that $i$ and $-i$ are two expressions for the same thing ; a priori they are not algebraically distinguishable. So writing $\sqrt{-1}$ itself is ambiguous, whether you're talking about branch cuts or not ; the expression "$\sqrt{-1}$" refers to a root of $x^2 + 1$, because when we use the symbol $\sqrt{-}$ in a context where it is well-defined, it is when we work with positive arguments and assume that the extracted root is positive. With negative arguments we have no such luck.

In other words, $\sqrt{-1}$ refers to a root of $x^2+1$, and if $i$ is such a root, then $-i$ is the other one. But if we call the roots $a$ and $b$ instead, all we know is $b = -a$, but nothing prevents us from reversing the roles.

So don't worry about it all that much.

Hope that helps,

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Thank you for the answer, Patrick. –  Lev Oct 10 '12 at 19:02