I am having trouble understanding how branch cuts work. For example, the function $f(z)= \sqrt{z}$ has a branch cut where you reject the negative real axis. But how do you define the output so that the function is 1-1 and onto? For example what is $f(1 + i)$? And $f(1-i)$?

Also does the function containing the branch cut have to be analytic?


1 Answer 1


It's easiest to understand the principal square root if you write the complex numbers in polar coordinates, that is, $r(\cos\theta+i\sin\theta)$ for some real $r\ge 0$ and real $\theta$ rather than $a+ib$ for real $a$ and $b$.

To find (the principal value of) $\sqrt z$, write $z=r(\cos\theta+i\sin\theta)$ with $\theta$ chosen in $(-\pi,\pi]$. Then $$\sqrt z = \sqrt r ( \cos\frac\theta 2 + i\sin\frac\theta 2 )$$

Thus, since $1+i = \sqrt 2(\cos\frac\pi4 + i\sin\frac\pi4)$ we get $\sqrt{1+i} = \sqrt[4]2(\cos\frac\pi8 + i\sin\frac\pi8)$ -- whose value in ordinary coordinates I don't care to create, but we can read immediately off the expression that it lies 22½° above the positive real axis and about 1.19 from the origin.

Note that the function thus defined happens to be 1-1 (though this is more by accident than by design), but it is not onto. Its value never lies to the left of the imaginary axis.

Finally: One usually speaks of branch cuts only when we're talking about analytic functions. There's nothing that formally prevents us from defining a non-analytic function that happens to have a branch cut, but this is generally not considered an especially interesting situation -- for non-analytic functions it is usually more fruitful to consider them as functions on $\mathbb R^2$ instead of on $\mathbb C$.

  • $\begingroup$ More advanced material is classified not "branch cuts" but "Riemann surfaces" ... you study how the branches fit together. $\endgroup$
    – GEdgar
    Apr 21, 2012 at 13:00

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