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If a is not in the image of f, and f is entire, then we know that there is a sequence in a domain whose value of function converges to a. Now the problem I faced is that even further we can choose a continuous curve p whose parameter is t in[0,1) such that p(t) goes to inf as t goes to 1, while f(p(t)) converges to a.

That's much stronger than image of entire function is dense in the plane .. Actually I faced two difficulties.

  1. How can we find a sequence whose value of functions converges to a? With its modulus goes to infinity?

  2. How can we connect these sequence to make a continuous curve whose value of function is not much varying between two points of consecutive points?

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The answer to your first question is easy. Since you know that the image of $f$ is dense in $\mathbb{C}$, you can find a sequence $(z_n)$ with $f(z_n) \to a$. If $(z_n)$ does not diverge to $\infty$, there exists a subsequence converging to some point $z^* \in \mathbb{C}$. Continuity then implies $f(z^*) = a$, contradicting the assumption that $a$ is not in the image of $f$.

The second question is in fact quite a bit harder, the result is a special case of Iversen's theorem. Just picking any such sequence it might be impossible to connect even a subsequence into a curve that works. If $a\ne \infty$, you can replace $f$ by $\tilde{f}(z)=\frac{1}{f(z)-a}$, which is again an entire function, and any curve on which $\tilde{f}$ tends to $\infty$ will be a curve on which $f$ tends to $a$. So you can assume that $a=\infty$. One standard proof of this result is via the Gross star theorem, which says that any local inverse near $w_0 = f(z_0)$ of an entire function can be continued along almost any ray emanating from $w_0$. If $R$ is any such ray, and $\gamma$ is the image of $R$ under the continuation of this local inverse, then $f \to \infty$ along $\gamma$, which in turn implies that $\gamma$ has to tend to $\infty$.

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