Proof of the solutions for complex conjugate roots

Proof

We work with complex solutions and take real and imaginary parts.

1. Conjugate root. For $h \in \C[t]$, let $\bar h$ be $h$ with each coefficient replaced by its complex conjugate. Conjugation respects sums and products of complex numbers, so $\overline{h_1 h_2} = \bar h_1 \bar h_2$. Also $\bar p = p$, since $p$ has real coefficients.
2. Let $\lambda = \alpha + i \beta$. By the definition of multiplicity, $p = (t - \lambda)^m h$ for some $h \in \C[t]$. Conjugating gives $p = (t - \bar\lambda)^m \bar h$. So the multiplicity of $\bar\lambda$ is at least $m$.
3. The same argument with $\bar\lambda$ in place of $\lambda$ shows that the multiplicity of $\lambda$ is at least that of $\bar\lambda$. So $\alpha - i \beta = \bar\lambda$ is a root of multiplicity exactly $m$.
4. Complex functions. For $u = u_1 + i u_2$ with real $u_1, u_2 : \R \to \R$, let $u' = u_1' + i u_2'$ when both are differentiable. The product rule and linearity with complex constants hold, by applying the real rules to real and imaginary parts.
5. For $\mu = a + i b$ with $a, b \in \R$, let $E_\mu(x) = e^{\mu x}$. By the product of exponentials and Euler's formula, $E_\mu(x) = e^{a x}\left(\cos b x + i \sin b x\right)$.
6. By the chain rule and the derivatives of exp, cos and sin, $E_\mu' = e^{a x}\left(a \cos b x - b \sin b x\right) + i\, e^{a x}\left(a \sin b x + b \cos b x\right)$. Expanding $(a + i b)(\cos b x + i \sin b x)$ shows that this is $\mu E_\mu$.
7. Also $E_\mu E_\nu = E_{\mu + \nu}$, and $E_\mu(x) E_{-\mu}(x) = E_0(x) = 1$. So $E_\mu(x) \ne 0$ for each $x$.
8. Let $D$ be differentiation of infinitely differentiable functions $\R \to \C$. For $q(t) = \sum_j b_j t^j \in \C[t]$, let $q(D) u = \sum_j b_j u^{(j)}$. As $D^j$ is linear and $D^j D^l = D^{j+l}$, we get $q_1(D)\left(q_2(D) u\right) = (q_1 q_2)(D) u$.
9. By steps 4 and 6, $(D - \mu)\left(x^k E_\mu\right) = k x^{k-1} E_\mu$ for $k \ge 0$. Applying this $k + 1$ times gives $(D - \mu)^{j}\left(x^k E_\mu\right) = 0$ for each $j > k$.
10. Solutions. Let $0 \le k \le m - 1$. By steps 2, 8 and 9, $p(D)\left(x^k E_\lambda\right) = h(D)\left((D - \lambda)^m \left(x^k E_\lambda\right)\right) = 0$.
11. By step 5, $x^k E_\lambda = u_k + i v_k$, where $u_k = x^k e^{\alpha x} \cos \beta x$ and $v_k = x^k e^{\alpha x} \sin \beta x$ are real.
12. The coefficients of $p$ are real. So $p(D) u_k$ and $p(D) v_k$ are real, and $0 = p(D)(u_k + i v_k) = p(D) u_k + i\, p(D) v_k$. Hence $p(D) u_k = p(D) v_k = 0$.
13. So the $2m$ functions $u_k, v_k$ are real solutions of the equation with constant coefficients on $\R$, with characteristic polynomial $p$.
14. Independence. Let $a_k, b_k \in \R$ with $\sum_{k=0}^{m-1} \left(a_k u_k + b_k v_k\right) = 0$ on $\R$.
15. The cosine is even and the sine is odd, as their definitions show. So $E_{\bar\lambda}(x) = e^{\alpha x}\left(\cos \beta x - i \sin \beta x\right)$. Hence $u_k = x^k \left(E_\lambda + E_{\bar\lambda}\right)/2$ and $v_k = x^k \left(E_\lambda - E_{\bar\lambda}\right)/(2i)$.
16. Substituting, $Q_1(x) E_\lambda(x) + Q_2(x) E_{\bar\lambda}(x) = 0$ for all real $x$. Here $Q_1(t) = \sum_k \frac{a_k - i b_k}{2} t^k$ and $Q_2(t) = \sum_k \frac{a_k + i b_k}{2} t^k$.
17. Multiply by $E_{-\bar\lambda}$ and use step 7: $Q_1(x) E_\gamma(x) + Q_2(x) = 0$, where $\gamma = \lambda - \bar\lambda = 2 i \beta \ne 0$.
18. Suppose $Q_1 \ne 0$, with degree $d$ and leading coefficient $c$. By step 6 and the product rule, $D\left(Q(x) E_\gamma\right) = \left(Q' + \gamma Q\right)(x)\, E_\gamma$ for each polynomial $Q$.
19. If $Q \ne 0$, then $Q' + \gamma Q$ has the same degree and leading coefficient multiplied by $\gamma \ne 0$, by the derivative of a polynomial.
20. Let $e$ be one more than the degree of $Q_2$, or $e = 0$ if $Q_2 = 0$. Differentiating step 17 $e$ times gives $R(x) E_\gamma(x) = 0$ for all real $x$, where $R$ has degree $d$ and leading coefficient $\gamma^e c \ne 0$.
21. By step 7, $R(x) = 0$ for all real $x$. A nonzero polynomial over $\C$ has finitely many roots, a contradiction. So $Q_1 = 0$.
22. Then $Q_2(x) = 0$ for all real $x$, so $Q_2 = 0$ by the same fact. Thus $a_k - i b_k = 0$ for each $k$, so $a_k = b_k = 0$.
23. Hence the $2m$ real solutions are linearly independent on $\R$.