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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? –  Chris Eagle Jul 29 '12 at 19:01

1 Answer 1

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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Thank you for your answer. For a bit of clarification: is log(z) holomorphic where x is positive because of the standard problem of taking the log of a negative number (and if yes, why does only x have to be positive), or is it different with complex numbers? Also, how did you know it jumps to 2πi? Excuse the extra question, I'm trying to teach myself complex analysis, and while my book provides explanations, there isn't much in the way of detailed examples, so I don't actually see how the complex logarithm works. –  JKH Jul 29 '12 at 19:23
    
This "jumping" is a very common convention, but the log function itself does not really "know" this convention. E.g., the power series expansion for $\log z$ at $z_o=-4+3i$ has radius of convergence $5$, not $3$, the latter being the distance to the supposed "slit" or "jump" along the negative real axis. –  paul garrett Jul 29 '12 at 19:58
    
@JKH: $\log(z)$ is supposed to be a (complex) number $w$ such that $e^{w} = z$. The problem is that $e^{2 \pi i} = 1$, so there is not a unique solution: you can add $2 \pi i$ to any solution and get another one. If you think of $z$ moving counterclockwise around the unit circle starting at $1$, a corresponding $w$ with $e^w = z$ will start at $0$ and move continuously up the imaginary axis. After one complete revolution $z$ is back at $1$, but $w$ is at $2 \pi i$. –  Robert Israel Jul 29 '12 at 22:00
    
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 $\infty$. –  Robert Israel Jul 29 '12 at 22:06
    
Thank you, your answer is helpful. –  JKH Aug 3 '12 at 20:00

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