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1) Let $a>0.$ Prove that $p(z)=z^3+3a^2z-1$ has a simple zero on $(0,1)$ and it doesn't have more real roots.

2) Consider a meromorphic function $f:\mathbb C\to\mathbb C$ so that $|f(z)|\xrightarrow[|z|\to\infty]{}\infty.$ Prove that $f$ is a rational function.

__Attempts__:

1) I'd need to use Rouché's Theorem here, so I think I need to split it in two cases, when $|z|=0$ and $|z|=1.$ For first case let $|p(z)-1|=|z^3+3a^2z|\le|z^3|+3a^2|z|=0\le1\le|p(z)|+1,$ since $p(z)$ and $1$ don't have zeroes for $|z|=0,$ then by Rouché's Theorem $p(z)$ has no zeros for $|z|>0.$ For $|z|=1$ let $|p(z)-3a^2z|=|z^3-1|\le 1<3a^2=3a^2|z|,$ here's my problem, I think it should be $a>1$ so $|p(z)-3a^2z|\le 3a^2|z|\le |p(z)|+3a^2|z|,$ since $p(z)$ and $3a^2z$ don't have zeroes for $|z|=1$ by Rouché's Theorem $p(z)$ has a single zero on $|z|<1.$ In conclusion $p(z)$ has only a root for $0<|z|<1$ and the zero is simple because $p'(z)=3z^2+3a^2\ne0.$

2) I think I can write $g(z)=p(z)f(z)$ where $p(z)$ is a monic polynomial whose roots are the poles of $f,$ then $g(z)$ is a entire function and has convergent Taylor series, so $g(z)=\displaystyle\sum_{j=0}^\infty\frac{g^{(j)}(0)}{j!}z^j,$ so the only thing I need to prove is that $g$ is a polynomial, but I don't see how to use the info. of the limit.

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