Let $w=w(z_1,z_2)$ be any polynomial on $\mathbb{C}^2$, set $T(z)=w(z,\bar{z})$, show that the image $T(\mathbb{C})\subset \mathbb{C}$ is closed.
The image of a polynomial map is not generally closed. For example, See here. What my try at first is to show the graph of $T$ is closed, and $T$ is continuous to infinity, and then the projection map is a closed map, Projection map being a closed map. However, the continuity is false. Could anyone give me some hints? Thanks very much!
This is false.
Let $P(z,w) = z + w + z^2 - w^2$
Then
$$ \begin{align*} P(z,\bar{z}) &= z+\bar{z} + (z+\bar{z})(z - \bar{z})\\ &= 2\textrm{Re}(z) + 4\textrm{Re}(z)\textrm{Im}(z)i \end{align*} $$
Letting $z = x+iy$ we have
$f(x,y) = P(x+iy, x-iy) = 2x + 4xyi$
The image of this map consists of the entire complex plane except for the positive and negative imaginary axis.
If $a \neq 0$ and $b \in \mathbb{R}$, then $f(\frac{1}{2}a, \frac{1}{2}b) = a + bi$. Hence every point not on the imaginary axis is in the image.
If $b \in \mathbb{R}$ then $f(x,y) = bi$ only has a solution if $x = 0$, but $f(0,y) = 0+0i$ for all $y$. Hence the only point on the imaginary axis in the image is $0+0i$.
As a counterpoint to @Steven counterexample, let's prove that if $w$ nonconstant, has no mixed terms so it is $c+\sum_{k=1}^na_kz_1^k+\sum_{k=1}^mb_kz_2^k$ and if ether $m \ne n$ or if $m =n \ne 0$ and $|a_n| \ne |b_n|$ then $T(z, \bar z)=w(z, \bar z)$ is surjective so $T(\mathbb C)=\mathbb C$
By using $T-a$ which is of the same form, it is enough to prove that $T$ has at least one zero, and actually, we will prove that it has at least $n$ isolated zeroes.
With a bit more care by analyzing the Sylvester resultant of $T(z, \bar z)=\sum p_k(z)\bar z^k, \overline {T(z, \bar z)}=\sum q_k(z)\bar z^k$ when regarded as polynomials in $\bar z$ with coefficients in $\mathbb C[[z]]$ which vanishes precisely when $T$ vanishes, one can easily show that if $m <n$ all the roots of $T$ are among (but not necesarily all) the roots of a regular (analytic) polynomial in $z$ of degree precisely $n^2$ and hence they are isolated and at most $n^2$, while looking at $T_1(z, \bar z)=\frac{\bar a_n T-b_n \bar T}{|a_n|^2-|b_n|^2}$ which is easily seen to have same roots as $T$ and is in the case $m<n$ shows the result (at most $n^2$ isolated zeroes) for $n=m, |a_n| \ne |b_n|$ too.
Wlog we can assume either $n >m$ or $n=m, |a_n|>|b_n|$ so (letting $b_k=0, n < k \le m$ in the first case) consider the quantity:
$$R=\max (1, \frac{|c|+\sum_{k=1}^n(|a_k|+|b_k|)}{|a_n|-|b_n|})$$
Consider $g(re^{i\theta})=b_ne^{-in\theta}+a_ne^{in\theta}$ and note that $|g(re^{i\theta})| \ge (|a_n|-|b_n|)r^n>0$. But if $r>R \ge 1$ we have that
$|g(re^{i\theta})-T(re^{i\theta})| \le |c|+\sum_{k=1}^{n-1}(|a_k|+|b_k|)r^k \le (|c|+\sum_{k=1}^{n-1}(|a_k|+|b_k|)r^{n-1}$
hence by our choice of $R$ and the fact that $r >R$ we get
$$|g(re^{i\theta})-T(re^{i\theta})|<(|a_n|-|b_n|)r^n \le |g(e^{i\theta)})$$
So $T$ doesn't vanish on $C_r$, the positively oriented circle of radius $r$, and with $d$ denoting the topological degree, we have that $d(T, C_r)=d(g,C_r)=d(h, C_r)=d(h, C_1)$ where $h(z)=a_nz^n-b_n\bar z^n$ (and using the obvious fact that $h$ doesn't vanish on the annulus $1<|z|<r$ so $d(h, C_1)=d(h, C_r)$)
But now on $C_1$ we have that $h(z)=a_nz^n(1-b_n\bar z^{2n}/a_n)$ and $\Re (1-b_n\bar z^{2n}/a_n)>0$ which means that $d(1-b_n\bar z^{2n}/a_n, C_1)=0$,
hence $d(h, C_1)=d(a_nz^n, C_1)+d(1-b_n\bar z^{2n}/a_n, C_1)=n$ so $d(T, C_r)=n$ which means that $T$ has at least $n$ isolated zeroes with positive multiplicity inside $C_r$ since the sum of the multiplicities of the isolated zeroes adds to $n$ by the definition of the topological degree, so we are done and have shown that $T$ is surjective in this case.
