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Find an upper bound for

$$\left\vert{\rm Log}(z+ai)\right\vert$$

where $z\in\left\{Re^{i\theta}\,\colon 0\le\theta\le\pi, R>0 {\rm , R \,fixed}\right\}$, $a>0$ and ${\rm Log}(w) = \log{\vert w\vert} + i{\rm Arg}(w)$.


So I started by following this method:

\begin{align} \left\vert{\rm Log}(z+ai)\right\vert &= \left\vert\log{|z+ai|} + i{\rm Arg}(z+ai)\right\vert \\ &\le \left\vert\log{|z+ai|}\right\vert + \left\vert i{\rm Arg}(z+ai)\right\vert \\ &= \left\vert\log{|z+ai|}\right\vert + {\rm Arg}(z+ai) \\ &\le \left\vert\log{(|z|+a)}\right\vert + {\rm Arg}(z+ai) \\ &= \left\vert\log{(R+a)}\right\vert + {\rm Arg}(z+ai) \end{align}

But i'm not really sure how to bound the argument. Is this the best way of going about it or is there a nicer way to hand this?

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  • $\begingroup$ If you fix a branch for the logarithm, the argument is already bounded. $\endgroup$ – Starfall May 8 '16 at 20:32
  • $\begingroup$ @Starfall What do you mean by fix a branch? $\endgroup$ – user2850514 May 8 '16 at 20:37
  • $\begingroup$ The complex logarithm is a multivalued function, furthermore even if we restrict it to be single valued there is a discontinuity when we loop around the origin (the argument increases by $ 2\pi $ even if we get back to the same point.) To alleviate this, we introduce a branch cut, i.e we cut the complex plane by a ray through the origin. This amounts to restricting the argument of the complex logarithm to a certain interval. $\endgroup$ – Starfall May 8 '16 at 20:40
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    $\begingroup$ For instance, define $ \log(z) = \log|z| + i \arg(z) $ where $ \arg(z) $ is taken to be in $ (-\pi, \pi] $. Now, it is pretty clear that your argument is bounded. $\endgroup$ – Starfall May 8 '16 at 20:56
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    $\begingroup$ You don't need to. The question asks you to find an upper bound, and you found an upper bound. $\endgroup$ – Starfall May 8 '16 at 21:02

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