A complex polynomial in $z$ and $\bar z$ contains no terms with $\bar z$ if and only if its $\bar z$-derivative is zero I am struggling with this exercise:

Let $p(x,\bar{z})=\sum a_{lm}z^l\cdot\bar{z}^m$ be a polynomial in $z$
  and $\bar{z}$ (so only finitely many $a_{lm}$ are non-zero). Show that
  p contains no terms with $m>0$ if and only if $\frac{\partial
 p}{\partial \bar{z}}=0$.

From earlier I have proved that $\frac{\partial z}{\partial z}=1,\frac{\partial \bar{z}}{\partial z}=0,\frac{\partial z}{\partial \bar{z}}=0, \frac{\partial \bar{z}}{\partial \bar{z}}=1$, and the leibniz-rule. So there is no problem in differntiating the polynomials. 
It is the $\leftarrow$ direction I'm struggling with. Because $\rightarrow$ is easy since if we have no $\bar{z}$ then by what I wrote earlier $\frac{\partial p}{\partial \bar{z}}$ must be zero.
Here is my attempt on the other way:
$\leftarrow$
Assume that $\frac{\partial p}{\partial \bar{z}}=0$, then $\sum _{l,m} a_{lm} z^l m \bar{z}^{m-1} =0$.
I think what I must do is put in a smart value of z so that all the sums are zero expept one, and hence in that sum m or $a_{lm}$ must be zero. But how do I find these z value for every m and l?
Contrapositively
I could do a contrapositive proof, but then I get the same problem:
Assume that there is atleast one m>0 with $a_{lm}>0$. Then I must show that $\sum _{l,m} a_{lm} z^l m \bar{z}^{m-1}$ has atleast one point where it is not zero. Here I only need to find one point not many points as in the last one, but how do I do it?
 A: Your idea of evaluating the polynomial at "special values" is a good one, but it is hard to choose appropriate values when the coefficients are arbitrary. 
Here's another idea, along the same lines: note that if $\frac{\partial}{\partial \bar z}p(z,\bar z)$ is identically zero, then all subsequent derivatives (with respect to either $\bar z$ or $z$) are $0$. On the other hand, you can explicitly write down subsequent derivates as polynomials. 
By systematically differentiating $p(z,\bar z)$ with respect to both $z$ and $\bar z$, you can isolate certain coefficients as constant terms. If $a_{lm}$, $m>0$, is the only constant term of some higher derivative of $p(z,\bar{z})$, it should be clear how to show that $a_{lm}=0$. 
A: Your calculation of the  derivative $\frac{\partial p}{\partial \bar{z}}$ is correct. But it is a polynomial in the variables $z$ and $\bar{z}$ and it is zero as a polynomial, not as a number. Moreover this polynomial is identically zero if it is zero on the unit circle ($z\bar{z}=1$). This means that all the coefficients (if I replace $\bar{z}$ with $z^{-1}$) of the polynomial disappear. But now your partial derivative looks like 
$\Sigma _{l,m} a_{lm}*z^l*m*z^{1-m}=\Sigma _{l,m} a_{lm}*m*z^{1+l-m}$. Since this is a polynomial (and not a Laurent polynomial) the exponents $1+l-m$ all have to be positive, hence the desired result. 
A: This can be a bit challenging for the one not familiar with techniques used when dealing with differentials.
Use these :
$$\frac{\partial^{l+m}}{\partial z^l\overline{z}^m}(z^j\cdot\overline{z}^k)=\left(\prod_{I=0}^{l-1}(j-I)\right)\cdot\left(\prod_{J=0}^{m-1}(k-J)\right)z^{j-l}\cdot\overline{z}^{k-m}$$
if $l\leq j,\,m\leq  k,$
equal to 0 if either $l>j$ or $m> k$
and
$$\frac{\partial^{l+m}}{\partial z^l\overline{z}^m}(z^l\cdot\overline{z}^m)=l!\cdot m!.$$
Then the proof follows immediately.
A: Hint: Look at $z=\bar z = 1$ and notice the bounds of summation are
$$\sum_{l=0}^N \sum_{\mathbf{m=0}}^M \ldots$$
