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Problem 14 in Chapter 1.5 of Conway's "Course in Functional Analysis" asks me to let $\lambda = \text{Area measure on} \{ z\in \mathbb{C}\: |z| < 1\}$ and consider the vectors $1,z,z^2,...$. If these vectors are normalized, as $e_n = \frac{z^n}{||z^n||}, n \geq 0$, then I am asked to determine if this is a basis for $L^2(\lambda)$.

I was realizing though, if you write the complex number $z$ in polar form, as $z=r e^{i\theta}$, then $e_n = e^{in\theta}$ seems to be the basis for trigonometric polynomials, and also the basis for a Fourier series.

I don't have much experience with complex numbers, so perhaps this is obvious, or perhaps I am missing something.

My question: what is the difference between:

  1. The polynomials of the form $\sum_{k=0}^\infty a_k z^k$ where $z \in \mathbb{C}$
  2. Trigonometric polynomials written as $\sum_{k=0}^\infty a_k e^{ik\theta}$

Are complex valued polynomials the same as trigonometric polynomials? Perhaps within a normalization factor?

I ask because I seem to recall learning that the trigonometric polynomials are dense in the space of continuous functions on an interval, and I would like to use that result to answer this question.

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Here's one reason why complex polynomials (which is what you are considering, since $z\in\mathbb{C}$) and trigonometric polynomials differ: If I consider a simple trigonometric polynomial $\cos(\theta)=(e^{i\theta}+e^{-i\theta})/2$, then as $\theta$ varies over $\mathbb{R}$, this has infinitely many zeroes, because of periodicity. On the other hand, complex polynomials factorize as $\prod_{i=1}^{n}(z-\alpha_{i})$, where $\alpha_{1},\ldots,\alpha_{n}\in\mathbb{C}$ (not necessarily distinct), and in particular, only the zero polynomial has infinitely many points in $\mathbb{C}$ for which it takes the value 0.

They are not unrelated, however. If I have a complex polynomial $\sum_{i=0}^{n}a_{i}z^{i}$, then I get a trigonometric polynomial by restricting $z$ to $\{z\in\mathbb{C}:|z|=1\}=\{e^{i\theta}:\theta\in\mathbb{R}\}.$

I will add that throughout this answer, I'm supposing that you mean true polynomials with finite degree, rather than power series as you've written.

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    So, they are the same when $|z| = 1$, which is essentially the case of this problem, since we are considering only $z \in \mathbb{C}$ where $z<1$, and then normalizing by $|z|$. – 2017-01-29
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    I still wouldn't say this, since usually trigonometric polynomials (for example, $\cos$) involve $e^{-in\theta}$, but polynomials do not allow for $z^{-n}$. Perhaps a closer analogy would be between trigonometric polynomials and Laurent polynomials (which do allow negative powers of $z$). – 2017-01-30