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Let $f(z)=z^n+nwz$ be a complex function with $|w|=1$ and $n>1$ a natural number. Is this function one-to-one inside the unit circle ($|z|<1$)?

ATTEMPT I didn't have a lot of luck checking $f(z_1)=f(z_2)$ and trying to manipulate the equation to get $z_1=z_2$. The factor of $n$ really invites taking the derivative, and I know there's a theorem that states something along the lines of, if $f$ is analytic in $R$ and $f'<>0$ then $f$ is one-to-one in R, but the proof of this theorem that I found relies on complex integration, which we haven't learned yet.

I figure I could try and separate $f$ to real and imaginary components and find the derivative of those, and do some ad-hoc analysis, but is there an easier solution?

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Let $z_1\neq z_2$ in the unit disk. Then $f(z_2)-f(z_1)=(z_2-z_1)\left(\sum_{j=0}^{n-1}z_2^jz_1^{n-1-j}-n\omega\right).$ The modulus of $\sum_{j=0}^{n-1}z_2^jz_1^{n-1-j}$ is strictly lower than $n$.

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    I missed this. Thank you!2012-12-02
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Note that $|f'(z) - n w | = |n z^{n-1}| < n$ for $z$ in the disk. If $\Gamma$ is the line segment from $z_1$ to $z_2$, where $z_1 \ne z_2$ are in the disk, $|f(z_2) - f(z_1) - n w (z_2 - z_1)| = \left|\int_\Gamma (f'(z) - n w)\ dz \right| so $f(z_2) - f(z_1) \ne 0$.