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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 $cos6-sin6i$

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    $\begingroup$ If $z_0^{15}+i=0$ for some $z_0$ then the function is not defined for $z_0$. $\endgroup$ – drhab Jan 1 '15 at 14:55
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    $\begingroup$ There are fifteen numbers $z$ such that $z^{15}=-i$, taking roots is a multivalued operation $\endgroup$ – David Peterson Jan 1 '15 at 14:59
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    $\begingroup$ 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." $\endgroup$ – drhab Jan 1 '15 at 15:02
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    $\begingroup$ @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. $\endgroup$ – Timbuc Jan 1 '15 at 15:09
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    $\begingroup$ Thanks @Drhab. That can be a really fruitful discussion as this can be a blurry point for many, up to and including me. $\endgroup$ – Timbuc Jan 1 '15 at 15:20
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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.

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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.

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