A complex-line is linear map L : $\mathbb{C}$ $\rightarrow$ $\mathbb{C}^{n}$ i.e. it is given by a (n $\times$ 1) complex matrix.
I am given that an entire function f : $\mathbb{C}^{n}$ $\rightarrow$ $\mathbb{C}$ is bounded on infinitely many complex-lines through origin, then f must be constant.
I was thinking because all of the planes pass through origin, may be there is a small open ball around origin where the function is constant, which would make the entire function constant everywhere. But I don't know how to actually show it. Any help would be appreciated.
This is incorrect if $n>2$. Indeed, $f(z_1,\dots,z_n)=z_1$ is nonconstant and vanishes only every complex-line through the origin in $\{0\}\times\mathbb{C}^{n-1}$, and if $n>2$ there are uncountably many such lines.
If $n=2$, the result is true, and here's a sketch of the argument: first, show $f$ is constant on the union of all your lines on which you know it is bounded. Now, lines through the origin are parametrized by the projective space $\mathbb{CP}^1$, which is compact. So if you have infinitely many of them, they will accumulate at some line through the origin $L$. Now if you restrict $f$ to any line $L'$ which intersects $L$ somewhere other than the origin, it will be constant on a subset of $L'$ that has an accumulation point, and thus constant on all of $L'$.