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This seems very obvious and I am having a bit of trouble producing a formal proof.

sketch proof that the composition of two polynomials is a polynomial

Let $$p(z_1)=a_nz^n_1+a_{n-1}z^{n-1}_1+...+a_1z_1+a_0 \\ q(z_2)=b_nz^n_2+b_{n-1}z^{n-1}_2+...+b_1z_2+b_0$$ be two complex polynomials of degree $n$ where $a_n,..,a_0\in\mathbb{C}$ and $b_n,..,b_o\in\mathbb{C}$.

Now, \begin{align} (p\circ q)(z_2)&=p(q(z_2)) \ \ \ \ \ \text{(by definition)}\\ &=a_n(q(z_2))^n+a_{n-1}(q(z_2))^{n-1}+...+a_1(q(z_2))+a_0 \end{align} which is clearly a complex polynomial of degree $n^2$.

sketch proof that the composition of two rational functions is a rational function

A rational function is a quotient of polynomials.

Let $$a(z_1)=\frac{p(z_1)}{q(z_1)}, \ b(z_2)=\frac{p(z_2)}{q(z_2)}$$ Now, \begin{align} (a\circ b)(z_2)&=a(b(z_2)) \ \ \ \ \ \text{(by definition)} \\ &=\frac{p\left(\frac{p(z_2)}{q(z_2)}\right)}{q\left(\frac{p(z_2)}{q(z_2)}\right)} \\ &=\frac{a_n\left(\frac{p(z_2)}{q(z_2)}\right)^n+a_{n-1}\left(\frac{p(z_2)}{q(z_2)}\right)^{n-1}+...+a_1\left(\frac{p(z_2)}{q(z_2)}\right)+a_0}{b_n\left(\frac{p(z_2)}{q(z_2)}\right)^n+b_{n-1}\left(\frac{p(z_2)}{q(z_2)}\right)^{n-1}+...+b_1\left(\frac{p(z_2)}{q(z_2)}\right)+b_0} \\ \end{align} Notice that $\left(\frac{p(z_2)}{q(z_2)}\right)^i \ \ \ \ (i=n, n-1,..,0)$ is a polynomial as $$(f\circ g)(z_2)=f(g(z_2))=\left(\frac{p(z_2)}{q(z_2)}\right)^i$$ where $$f(x)=x^i, \ \ g(z_2)=\left(\frac{p(z_2)}{q(z_2)}\right)$$ are both polynomials. Hence $(a\circ b)(z_2)$ is a rational function as it is the quotient of polynomials.

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Remark for your proof of composition of polynomials is a polynomial:

  • perhaps you should work with two arbitrary polynomials of degree $m$ and $n$ to have generality.

Remark for your proof of composition of composition rational functions is a rational function:

  • you wrote $a(z_1)=\frac{p(z_1)}{q(z_1)}$ and $b(z_2) = \frac{p(z_2)}{q(z_2)}$ which means $a$ and $b$ seems to be the same function.

  • $\left( \frac{p(z_1)}{q(z_2)}\right)^i$ is a rational function rather than a polynomial.

Guide for proof of composition of polynomials is a polynomial:

  • First prove that products of two polynomials is a polynomial. Once you can do that, we have that suppose $p$ is a polynomial, then $p(x)^i$ is a polynomials by mathematical induction.
  • Prove that the set of polynomials is closed under scalar multiplication.
  • Prove that the set of polynomials is closed under addition.
  • With those lemmas (tools), I believe now you can prove that composition of polynomials is a polynomials. (remember to use arbitrary polynomials of degree $m$ and degree $n$.)

Guide for proof of composition of rational functions is a rational function:

  • First prove that products of two rational function is a rational function. Once you can do that, we have that suppose $h$ is a rational function, then $h(x)^i$ is a rational function by mathematical induction.
  • Prove that the set of rational functions is closed under scalar multiplication.
  • Prove that the set of rational function is closed under addition.
  • Prove that the set of rational function is closed under division.
  • With those lemmas (tools), I believe now you can prove that composition of rational functions is a rational function. Let $a(z)= \frac{p(z)}{q(z)}$ and $b(z) = \frac{r(z)}{s(z)}$ and use those tools that you have verified.
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  • $\begingroup$ Please elaborate. Regarding 1st, could construe your steps as: First, take two arb. poly. (with given values for degree $m,n$); but that seems useless as a general value need be there. Also, why need prove $i$-th power too, only the first part seems necessary. 2nd & 3rd steps are not clear about their significance to me. Also, how they act as lemmas is unclear still. Regarding the 2nd proof, the 1st step is unclear & surely need elaboration. Also, why need prove the $i$-th power is again unclear. The rest three steps are unable to be understood by me in terms of their importance. $\endgroup$ – jiten Jul 9 at 14:00
  • $\begingroup$ It is a unique answer, & request help by elaborating your answer, as feel would help me understand the proof process for the same a lot.If possible, can chat too in some new room (better if created by you). I stated it as unique answer, as the other (similar) questions ask in terms of abstract algebra terminology. But, this question asks in basic lingo, the one I wanted to ask myself. $\endgroup$ – jiten Jul 9 at 14:08
  • $\begingroup$ I am hoping you would help me, as the answer fits my needs; just need to understand by elaboration. $\endgroup$ – jiten Jul 9 at 14:45
  • $\begingroup$ Let $f=x^{1000}$ and $g=x+1$, how would you prove that $f\circ g$ is a polynomial? it is a consequence of the closure under multiplication, but are you going to repeat the argument each time? Hence, given a problem, think through what tools are needed to solve a problem and build the tools rather than repeating the same work multiple times. $\endgroup$ – Siong Thye Goh Jul 9 at 15:54
  • $\begingroup$ Can we discuss further in chat room? Have created one at: chat.stackexchange.com/rooms/95949/polynomial-composition . $\endgroup$ – jiten Jul 10 at 0:37

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