I am looking for entire functions $f(z)$ which satisfies the following conditions:

$$|zf(z)-3+e^{2z}|\leq 4+|z|$$ for all $z\in \mathbb{C}$

I proceed like below: Let $g(z)=zf(z)-3+e^{2z}$, then $g(z)$ is entire and $|g(z)|\leq 4+|z|$.and

$|g(z)|\leq |z|$ for $|z|>R$ for sufficiently large $R$.

Then by Cauchy estimates $g(z)$ is a polynomial of degree less or equal to $1$ for $|z|>R$. $g(z)=a+bz= zf(z)-3+e^{2z}$ imples $$f(z)=a/z+b+3/z -e^{2z}/z$$

Which is not entire. I can choose $a=-3$ to get rid of $a/z+3/z$ but still I do not get entire function. What can be said for $|z|\leq R$. I guess I am missing some ideas.

Suggestions please!


1 Answer 1


You have:

$$a + bz = zf(z) - 3 + e^{2z}$$

Plugging in $z = 0$ gives:

$$a = -3 + 1 = -2$$

So you get

$$-2 + bz = zf(z) - 3 + e^{2z}$$

And so:

$$f(z) = \frac{1 - e^{2z}}{z} + b$$

Now, this function has a removable discontinuity at $z = 0$, in particular if you set $f(0) = b$ you get an entire function.

  • $\begingroup$ Thanks , I did not realize I had to get the coeficients. $\endgroup$
    – mp100
    Commented Jan 8, 2015 at 20:25

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