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${{{z^4} + z - 2i} \over {{z^{15}} + i}}$ is the function continuous at every point in the complex plane?

I tried to do like this but it is right? $$\eqalign{ & {z^{15}} + i = 0 \cr & {z^{15}} = - i \cr}$$ $$\eqalign{ & z = - {i^{1/15}} \cr & z = 1\angle {{ - 90} \over {15}} \cr & z = 1\angle - 6 \cr & z = \cos 6 - \sin 6i \cr} $$

So the funtion discontinuous at $cos6-sin6i$

razercreed
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  • Is that what you mean? It's not even defined for all points. – user2345215 Jan 01 '15 at 14:54
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    If $z_0^{15}+i=0$ for some $z_0$ then the function is not defined for $z_0$. – drhab Jan 01 '15 at 14:55
  • @drhab so my answer is wrong? – razercreed Jan 01 '15 at 14:57
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    There are fifteen numbers $z$ such that $z^{15}=-i$, taking roots is a multivalued operation – David P Jan 01 '15 at 14:59
  • @razercreed: the argument of complex numbers should be expessed in radians. – Bernard Jan 01 '15 at 15:00
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    The function can only be discontinuous at $z_0$ if it is defined at $z_0$. The answer to the (weird) question should be: "no, the function is not continuous at every point in the complex plane simply because it is not defined at every point in the complex plane." A less weird question would be: "is the function continuous (on its domain in $\mathbb C$)? Then the right answer is: "yes, it is." – drhab Jan 01 '15 at 15:02
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    @drhab, I'm not sure your definition of "discontinuous at some point" would be agreed on by all. For one, I'd say a function is discontinuous at $;z_0;$ if it is defined in some (left, right or two-sided) neighbourhood of $;z_0;$ but either it isn't defined at $;z_0;$ or else at that point the function doesn't fulfill the condition of continuity. – Timbuc Jan 01 '15 at 15:09
  • The point is this is just wrong and confusing question and I'm voting to close it. – user2345215 Jan 01 '15 at 15:11
  • @user2345215 Even accepting drhab's definition of non-continuity, I can't see how this function is wrong or confusing: it asks whetehr the given function is continuous at the plane. Nothing wrong or confusing there. – Timbuc Jan 01 '15 at 15:14
  • Note that some discontinuities may be removable. – Mark Bennet Jan 01 '15 at 15:19
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    @Timbuc Not by all? So be it then. Maybe I will pose a question about this myself once. I will let you know then. – drhab Jan 01 '15 at 15:20
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    Thanks @Drhab. That can be a really fruitful discussion as this can be a blurry point for many, up to and including me. – Timbuc Jan 01 '15 at 15:20
  • @MarkBennet Removable discontinuity points still are discontinuity points. – Timbuc Jan 01 '15 at 15:21
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    @Timbuc That is one way of looking at it. It is better to say that the function is not defined at those points, and here there are definitely more discontinuities that can be dealt with by any factors of the numerator. The range and domain of the function are not given here so the question is not wholly specified. – Mark Bennet Jan 01 '15 at 15:25
  • @MarkBennet I agree with you mostly (that it depends...), but the question asks about continuiyt in the plane . That means, imo, that the OP is asking about the maximal possible definition domain of the function, and then dhrab's point kicks in. – Timbuc Jan 01 '15 at 15:29
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    @Timbuc Posted a question – drhab Jan 01 '15 at 15:40

2 Answers2

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$$z^{15}=-i=e^{\frac{3\pi i}2+2k\pi i}\implies z_k:=e^{\frac{\pi i}{30}\left(3+4k\right)}\;,\;\;k=0,1,2,...,14$$

At all the above $\;15\;$ values of $\;z\;$ the functions's denominator vanishes and thus the function cannot be continuous there.

Timbuc
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The answer to this somehow weird question is: "no, the function is not continuous at every point in the complex plane simply because it is not defined at every point in the complex plane."

A less weird question would be: "is the function continuous (on its domain in $\mathbb C$)?"

The answer to that question is yes. It is continuous at every $z_0\in\mathbb C$ for wich the function is defined.

drhab
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