# Complex integral of the modulus of a polynomial

In Rudin's Functional Analysis book there is this exercise (Chapter 8, Exercise 1b):

Let $P$ be a polynomial in $\mathbb{C}^n$. Prove that if $$\int_{T^n} |P| = 0,$$ where $T^n$ is the torus in $\mathbb{C}^n$: $T^n = \{(e^{i t_1},\dots,e^{i t_n})\colon t_i \in \mathbb{R},\ i = 1,\dots,n\}$, then $P$ is identically $0$. (Hint: Compute $\int_{T^n} |P|^2$).

There is something that I don't understand here. Let $n = 1$ and $P(z) = 1$. Then $\int_{T^1} |P| = 0$, however $P \neq 0$. What am I doing wrong?

Edit: I think I know where is my mistake. The integral $\int_{T^1} |P|$ is not complex integral, but rather (is it right?) $$\int_{T^1} |P| = (2\pi)^{-1} \int_0^{2\pi} |P(e^{i t})| dt.$$ Then, however, what's the point of the exercise? If the above definition is correct, then |P| is non-negative and hence $\int_{T^1} |P| = 0$ if and only if $P = 0$.

What else am I missing?

• Just to be 100% clear is $P$ a polynomial in one variable or more? – N8tron May 31 '18 at 17:37
• I just see two ways you could state this $P \in \mathbb{C}[x_1,\ldots x_n]$ or $P \in \mathbb{C}^n[X]$ – N8tron May 31 '18 at 17:38
• As far as I understand if $n=1$, then $P$ is a complex polynomial in one (complex) variable. If $n > 1$ then $P(z) = \sum c_\alpha z^\alpha$, where $\alpha$ is a multi-index and $z \in \mathbb{C}^n$. – xen May 31 '18 at 17:45
• So both coefficients $c_\alpha$ are vectors in $\mathbb{C}^n$ and and you have multiple variables $z^\alpha=\prod x_i^{\alpha_i}$? – N8tron May 31 '18 at 17:49
• @N8tron $c_\alpha$ is complex number and $z^\alpha = \prod z_i^{\alpha_i}$, where $z_i \in \mathbb{C}$. But even in the case $n=1$ I don't understand what is going on. It seems to me that the exercise is trivial because we basically integrate nonnegative function, hence the integral should be zero if and only if the function is zero. – xen May 31 '18 at 17:57