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Solving sin(π§)=2 (Trigonometric Equations with Complex Numbers)
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"# Solving sin(z)=2 (Trigonometric Equations with Complex Numbers)\n", | |
"\n", | |
"## Step 1: Writing sin(z) in terms of z\n", | |
"\n", | |
"Start with Euler's formula:\n", | |
"\n", | |
"$$\n", | |
"e^{i\\theta} = \\cos\\theta + i\\sin\\theta\n", | |
"$$\n", | |
"\n", | |
"Put $\\theta = z$:\n", | |
"\n", | |
"$$\n", | |
"e^{iz} = \\cos{z} + i\\sin{z}\n", | |
"$$\n", | |
"\n", | |
"Put $\\theta = -z$:\n", | |
"\n", | |
"$$\n", | |
"\\begin{aligned}\n", | |
"e^{-iz} &= \\cos(-z) + i\\sin(-z) \\\\\n", | |
"&= \\cos{z} - i\\sin{z}\n", | |
"\\end{aligned}\n", | |
"$$\n", | |
"\n", | |
"Subtract the 2 formulae above and make $\\sin{z}$ the subject:\n", | |
"\n", | |
"$$\n", | |
"\\begin{array}{rrl}\n", | |
" & e^{iz} \\!\\!\\!\\!\\! &= \\cos{z} + i\\sin{z} \\\\\n", | |
"- & e^{-iz} \\!\\!\\!\\!\\! &= \\cos{z} - i\\sin{z} \\\\\n", | |
"\\hline\n", | |
"& e^{iz} - e^{-iz} \\!\\!\\!\\!\\! &= 2i\\sin{z} \\\\\n", | |
"& \\implies \\sin{z} \\!\\!\\!\\!\\! &= \\displaystyle\\frac{e^{iz} - e^{-iz}}{2i}\n", | |
"\\end{array}\n", | |
"$$\n", | |
"\n", | |
"## Step 2: Solving the equation\n", | |
"\n", | |
"Substitute the result from above:\n", | |
"\n", | |
"$$\n", | |
"\\begin{aligned}\n", | |
"\\sin{z} &= 2 \\\\\n", | |
"\\displaystyle\\frac{e^{iz} - e^{-iz}}{2i} &= 2 \\\\\n", | |
"e^{iz} - e^{-iz} &= 4i \\\\\n", | |
"(e^{iz})^2 - 1 &= 4i(e^{iz}) \\\\\n", | |
"(e^{iz})^2 - 4i(e^{iz}) - 1 &= 0 \\\\\n", | |
"\\end{aligned}\n", | |
"$$\n", | |
"\n", | |
"Apply quadratic formula by substituting $u = e^{iz}$:\n", | |
"\n", | |
"$$\n", | |
"\\begin{aligned}\n", | |
"e^{iz} &= \\displaystyle\\frac{-(-4i) \\pm \\sqrt{(-4i)^2 - 4(1)(-1)}}{2(1)} \\\\\n", | |
"&= (2 \\pm \\sqrt{3})i \\\\\n", | |
"iz &= \\ln((2 \\pm \\sqrt{3})i) \\\\\n", | |
"&= \\ln(2 \\pm \\sqrt{3}) + \\ln(i) \\\\\n", | |
"iz \\cdot (-i) &= \\left[ \\ln(2 \\pm \\sqrt{3}) + \\ln(i) \\right] \\cdot (-i) \\\\\n", | |
"z &= -\\ln(2 \\pm \\sqrt{3})i - \\ln(i)i\n", | |
"\\end{aligned}\n", | |
"$$\n", | |
"\n", | |
"## Step 3: Calculating ln(i)\n", | |
"\n", | |
"Calculate $\\ln(i)$ by writing $i$ in exponential form:\n", | |
"\n", | |
"$$\n", | |
"\\begin{aligned}\n", | |
"i &= \\cos\\theta + i\\sin\\theta \\\\\n", | |
"&= \\cos(\\pi/2 + 2k\\pi) + i\\sin(\\pi/2 + 2k\\pi) & \\forall k \\in \\mathbb{Z} \\\\\n", | |
"&= e^{(\\pi/2 + 2k\\pi)i} \\\\\n", | |
"\\ln(i) &= \\ln(e^{(\\pi/2 + 2k\\pi)i}) \\\\\n", | |
"&= (\\pi/2 + 2k\\pi)i\n", | |
"\\end{aligned}\n", | |
"$$\n", | |
"\n", | |
"## Step 4: Rewriting the solution\n", | |
"\n", | |
"Rewrite the solution:\n", | |
"\n", | |
"$$\n", | |
"\\begin{aligned}\n", | |
"z &= -\\ln(2 \\pm \\sqrt{3})i - \\ln(i)i \\\\\n", | |
"&= -\\ln(2 \\pm \\sqrt{3})i - [(\\pi/2 + 2k\\pi)i]i \\\\\n", | |
"&= -\\ln(2 \\pm \\sqrt{3})i + (\\pi/2 + 2k\\pi) \\\\\n", | |
"&= (\\pi/2 + 2k\\pi) - \\ln(2 \\pm \\sqrt{3})i\n", | |
"\\end{aligned}\n", | |
"$$\n", | |
"\n", | |
"--------\n", | |
"\n", | |
"This gist was inspired by [Math for fun, sin(z)=2](https://youtu.be/3C_XD_cCeeI)." | |
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