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Let $f:\mathbb{C}\to\mathbb{C}$, $f(z)=z^n$. Then $f$ maps sectors to disk. Say, if $z=\frac{1}{2}\left[\cos(\pi/4)+i\sin(\pi/4)\right]$ and $n=4$ then $f(z)= \frac{1}{16}\left[\cos(\pi)+i\sin(\pi)\right]$. But what about the rest of the (I think three more) points, how are they mapped?

I tried this: $z=\frac{1}{2}\left[\cos(\pi/4+2k\pi)+i\sin(\pi/4+2k\pi)\right]$, then $f(z)= \frac{1}{16}\left[\cos(\pi+8k\pi)+i\sin(\pi+8k\pi)\right]$. Not much help.

Would appreciate some clarification. I'm sure I'm missing an important detail.

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    what three points are you referring to? I also assume you mean the (1/2) in parentheses outside... like $\frac{1}{2}(\cos(\pi/4) + i \sin(\pi/4))$? – 2017-01-08
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    Right, edited, thanks for pointing that out. $f$ is supposed to map $n$-th roots of a complex number in a sector of a disc to a whole disc. That's why if $n=4$ and $\left| z \right|=1$, we must have that all numbers in the first quadrant must be mapped to the entire unit disc. But I don't see how that happens, since any roots of any $z^n$ will be mapped to $z^n$. That is $z_j$ ($1\le j \le 4$) will be mapped to $z^n$. So I'm quite confused. – 2017-01-08
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    What is $z_j$? Is your concern that the modulus changes? – 2017-01-08
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    $z_j$ is a 4th root of unity in that case. I'd say that $f(z^n)=z$ maps sectors to disc, not vice versa, so I'm wondering if that is a typo in my source. – 2017-01-08
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    The fourth roots of unity will all be mapped to 1 by $z^4$ by definition – 2017-01-08
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    Yes, but the thing is that in my source it is written that $f(z)=z^n$ must map sectors to disc. Say, if $n=4$ then one of the sectors is 1/4-th of the disc of radius $|z|$ in the 1st quadrant, which should be mapped by $f$ to the disc of radius $|z|$. – 2017-01-08
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    Yes, $z^4$ maps a the sector of the unit disk in the first quadrant to the unit disk. It maps the part of the 1st quadrant outside the unit circle to the whole outside of the unit circle – 2017-01-08
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    But how does that happen if $f(z^{1/4})=z$, so four points get mapped to only one point? – 2017-01-08
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51392/discussion-between-spaceisdarkgreen-and-sequence). – 2017-01-08
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    It might help to try an parameterize a sector. Can you say what sector $S = \{re^{i\theta} : r \in [0,1],\, \theta \in [0, \pi/6]\}$ is described here (geometrically)? Can you figure out what $n$ needs to be, in this case, so that $f: z \mapsto z^n $ maps $S$ to the unit disc? – 2017-01-08
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    $S$ is $1/6$-th of a disc in the first quadrant, $n=6$. – 2017-01-08

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