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I need to determine where the following function is differentiable and holomorphic in $\mathbb C$:

$f(z)=(z-3)^i$

I have the derivative as $df/dz= i(z-3)^{-1+i}$. The answer in my book says f is differentiable and holomorphic on $\mathbb C$ where $y\neq0$ and $x>3$. I don't see where this comes from. wolframalpha plotted the derivative and I can see that the imaginary part has a vertical asymptote at 0, but I can't see why f is not differentiable or holomorphic for $x\leq3$. How can I find the answer by looking at the function and its derivative?

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    How are you defining the complex exponential?2012-07-29

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The standard definition of $z^b$ for complex $b$ is $\exp(b \log z)$. Different branches of the logarithm will give you different branches of $z^b$. If you use the principal branch of the logarithm (which is what Wolfram Alpha uses, and apparently what your book uses), $\log z$ (and thus $z^b$) will be holomorphic in ${\mathbb C} \backslash (-\infty,0]$, and so $\log(z-3)$ and $(z-3)^b$ will be holomorphic in ${\mathbb C} \backslash (-\infty,3]$. As $z$ crosses the line $(-\infty, 3)$, $\log(z-3)$ jumps by $2 \pi i$ so $z^b$ changes by a factor of $\exp(2 \pi i b)$.

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    So if you want to define $\log(z)$ as a continuous function, you have a problem: it can't be both $0$ and $2 \pi i$ at the same point. Any version of $\log(z)$ must have a discontinuity at some point on the unit circle (and similarly on any circle surrounding the origin). The "principal branch" convention is to take this jump on the negative real axis. But you could do it on any other ray from $0$, or indeed on any curve from $0$ to complex $\infty$.2012-07-29