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  "body": "\n## Proof\n\nWe work with complex solutions and take real and imaginary parts.\n\n1. Conjugate root. For $h \\in \\C[t]$, let $\\bar h$ be $h$ with each coefficient replaced by its [[complex conjugate|0BJV24C4SY]]. 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.\n2. Let $\\lambda = \\alpha + i \\beta$. By the definition of [[multiplicity|03XVP8Q4GY]], $p = (t - \\lambda)^m h$ for some $h \\in \\C[t]$. Conjugating gives $p = (t - \\bar\\lambda)^m \\bar h$. So the multiplicity of $\\bar\\lambda$ is at least $m$.\n3. 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 $m$.\n4. 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|0V9A43H7WV]] and [[linearity|0M0TKFRX2A]] with complex constants hold, by applying the real rules to real and imaginary parts.\n5. For $\\mu = a + i b$ with $a, b \\in \\R$, let $E_\\mu(x) = e^{\\mu x}$. By the [[product of exponentials|0WM7DJCM6G]] and [[Euler's formula|04R3CY41VJ]], $E_\\mu(x) = e^{a x}\\left(\\cos b x + i \\sin b x\\right)$.\n6. By the [[chain rule|0RXX2PT1MQ]] and the derivatives of [[exp|0EX9GFJ0B3]], [[cos|0V2DV29EGT]] and [[sin|00R4MWW532]], $E_\\mu' = e^{a x}\\left(a \\cos b x - b \\sin b x\\right) + i\\, e^{a x}\\left(a \\sin b x + b \\cos b x\\right)$. Expanding $(a + i b)(\\cos b x + i \\sin b x)$ shows that this is $\\mu E_\\mu$.\n7. 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 $x$.\n8. 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 $q_1(D)\\left(q_2(D) u\\right) = (q_1 q_2)(D) u$.\n9. 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 $j \u003e k$.\n10. Solutions. Let $0 \\le k \\le m - 1$. By steps 2, 8 and 9, $p(D)\\left(x^k E_\\lambda\\right) = h(D)\\left((D - \\lambda)^m \\left(x^k E_\\lambda\\right)\\right) = 0$.\n11. 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.\n12. 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 $p(D) u_k = p(D) v_k = 0$.\n13. So the $2m$ functions $u_k, v_k$ are real solutions of the [[equation with constant coefficients|0A6PGTTP2A]] on $\\R$, with [[characteristic polynomial|0QDXMG46QW]] $p$.\n14. 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 $\\R$.\n15. The cosine is even and the sine is odd, as their [[definitions|0A7YDKRNR6]] 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 $v_k = x^k \\left(E_\\lambda - E_{\\bar\\lambda}\\right)/(2i)$.\n16. 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 $Q_2(t) = \\sum_k \\frac{a_k + i b_k}{2} t^k$.\n17. Multiply by $E_{-\\bar\\lambda}$ and use step 7: $Q_1(x) E_\\gamma(x) + Q_2(x) = 0$, where $\\gamma = \\lambda - \\bar\\lambda = 2 i \\beta \\ne 0$.\n18. 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 $Q$.\n19. 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|0MH1WK89PP]].\n20. 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 $\\gamma^e c \\ne 0$.\n21. By step 7, $R(x) = 0$ for all real $x$. A nonzero polynomial over $\\C$ [[has finitely many roots|0KJVQV3E0F]], a contradiction. So $Q_1 = 0$.\n22. 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 $a_k = b_k = 0$.\n23. Hence the $2m$ real solutions are [[linearly independent|0EVHJFTG9Z]] on $\\R$.\n\n## Idea\n\nOver $\\C$, the root $\\lambda$ of multiplicity $m$ gives the solutions $x^k e^{\\lambda x}$, as for real roots. Since the equation has real coefficients, the real and imaginary parts of a complex solution are real solutions. Independence reduces, through Euler's formula, to the independence of $x^k e^{\\lambda x}$ and $x^k e^{\\bar\\lambda x}$ with $\\lambda \\ne \\bar\\lambda$.\n",
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