(L17 ex. 3) Use Sage or sympy to show that its solution to \frac{\partial^2 Y}{\partial z^2} = i\omega Y, k_1\sin(z\sqrt{-i\omega}) + k_2\cos(z\sqrt{-i\omega}), is mathematically equivalent to Y(z) = c_1e^{z\sqrt{i\omega}} + c_2e^{-z\sqrt{i\omega}}.
\newcommand{\Bold}[1]{\mathbf{#1}}k_{1} \sin\left(\sqrt{-i \, \omega} z\right) + k_{2} \cos\left(\sqrt{-i \, \omega} z\right)
\newcommand{\Bold}[1]{\mathbf{#1}}k_{1} \sin\left(\sqrt{-i \, \omega} z\right) + k_{2} \cos\left(\sqrt{-i \, \omega} z\right)
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\newcommand{\Bold}[1]{\mathbf{#1}}c_{1} e^{\left(\sqrt{i \, \omega} z\right)} + c_{2} e^{\left(-\sqrt{i \, \omega} z\right)}
\newcommand{\Bold}[1]{\mathbf{#1}}c_{1} e^{\left(\sqrt{i \, \omega} z\right)} + c_{2} e^{\left(-\sqrt{i \, \omega} z\right)}
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\newcommand{\Bold}[1]{\mathbf{#1}}\left(k_{1} \sin\left(\sqrt{-i \, \omega} z\right) + k_{2} \cos\left(\sqrt{-i \, \omega} z\right), c_{1} e^{\left(\sqrt{i \, \omega} z\right)} + c_{2} e^{\left(-\sqrt{i \, \omega} z\right)}\right)
\newcommand{\Bold}[1]{\mathbf{#1}}\left(k_{1} \sin\left(\sqrt{-i \, \omega} z\right) + k_{2} \cos\left(\sqrt{-i \, \omega} z\right), c_{1} e^{\left(\sqrt{i \, \omega} z\right)} + c_{2} e^{\left(-\sqrt{i \, \omega} z\right)}\right)
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Finding the first few terms of the Taylor series for these expressions suggests that k_1 = (c_1 - c_1)/i and k_2 = c_1 + c_2. We substitute these into expression A.
\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{720} \, {\left(i \, c_{1} \omega^{3} + i \, c_{2} \omega^{3}\right)} z^{6} - \frac{1}{120} \, {\left(c_{1} \sqrt{i \, \omega} \omega^{2} - c_{2} \sqrt{i \, \omega} \omega^{2}\right)} z^{5} - \frac{1}{24} \, {\left(c_{1} \omega^{2} + c_{2} \omega^{2}\right)} z^{4} - \frac{1}{6} \, {\left(-i \, c_{1} \sqrt{i \, \omega} \omega + i \, c_{2} \sqrt{i \, \omega} \omega\right)} z^{3} + \frac{1}{2} \, {\left(i \, c_{1} \omega + i \, c_{2} \omega\right)} z^{2} + {\left(c_{1} \sqrt{i \, \omega} - c_{2} \sqrt{i \, \omega}\right)} z + c_{1} + c_{2}
\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{720} \, {\left(i \, c_{1} \omega^{3} + i \, c_{2} \omega^{3}\right)} z^{6} - \frac{1}{120} \, {\left(c_{1} \sqrt{i \, \omega} \omega^{2} - c_{2} \sqrt{i \, \omega} \omega^{2}\right)} z^{5} - \frac{1}{24} \, {\left(c_{1} \omega^{2} + c_{2} \omega^{2}\right)} z^{4} - \frac{1}{6} \, {\left(-i \, c_{1} \sqrt{i \, \omega} \omega + i \, c_{2} \sqrt{i \, \omega} \omega\right)} z^{3} + \frac{1}{2} \, {\left(i \, c_{1} \omega + i \, c_{2} \omega\right)} z^{2} + {\left(c_{1} \sqrt{i \, \omega} - c_{2} \sqrt{i \, \omega}\right)} z + c_{1} + c_{2}
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\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{720} i \, k_{2} \omega^{3} z^{6} - \frac{1}{120} \, k_{1} \sqrt{-i \, \omega} \omega^{2} z^{5} - \frac{1}{24} \, k_{2} \omega^{2} z^{4} + \frac{1}{6} i \, k_{1} \sqrt{-i \, \omega} \omega z^{3} + \frac{1}{2} i \, k_{2} \omega z^{2} + k_{1} \sqrt{-i \, \omega} z + k_{2}
\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{720} i \, k_{2} \omega^{3} z^{6} - \frac{1}{120} \, k_{1} \sqrt{-i \, \omega} \omega^{2} z^{5} - \frac{1}{24} \, k_{2} \omega^{2} z^{4} + \frac{1}{6} i \, k_{1} \sqrt{-i \, \omega} \omega z^{3} + \frac{1}{2} i \, k_{2} \omega z^{2} + k_{1} \sqrt{-i \, \omega} z + k_{2}
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\newcommand{\Bold}[1]{\mathbf{#1}}-i \, c_{1} \sin\left(\sqrt{-i \, \omega} z\right) + c_{1} \cos\left(\sqrt{-i \, \omega} z\right) + i \, c_{2} \sin\left(\sqrt{-i \, \omega} z\right) + c_{2} \cos\left(\sqrt{-i \, \omega} z\right)
\newcommand{\Bold}[1]{\mathbf{#1}}-i \, c_{1} \sin\left(\sqrt{-i \, \omega} z\right) + c_{1} \cos\left(\sqrt{-i \, \omega} z\right) + i \, c_{2} \sin\left(\sqrt{-i \, \omega} z\right) + c_{2} \cos\left(\sqrt{-i \, \omega} z\right)
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This is clearly equivalent to expression B by Euler's Formula.