// Numbas version: finer_feedback_settings
{"name": "Contour integral of a complex-valued function", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variable_groups": [], "variables": {"b1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1..6)", "description": "", "name": "b1"}, "c1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(2..5)", "description": "", "name": "c1"}, "a1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1..6)", "description": "", "name": "a1"}}, "ungrouped_variables": ["a1", "c1", "b1"], "question_groups": [{"pickingStrategy": "all-ordered", "questions": [], "name": "", "pickQuestions": 0}], "name": "Contour integral of a complex-valued function", "functions": {}, "showQuestionGroupNames": false, "parts": [{"scripts": {}, "gaps": [{"answer": "0", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "checkvariablenames": false, "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "type": "jme", "showCorrectAnswer": true, "marks": 1, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "<p>$z=0$ and $z=\\simplify{1/{c1}}$ are both outside $C$.</p>\n      <p>$I=$ [[0]].</p>", "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"answer": "{-2*a1}*pi*i", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "checkvariablenames": false, "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "type": "jme", "showCorrectAnswer": true, "marks": 1, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "<p>$z=0$ is inside $C$, but $z=\\simplify{1/{c1}}$ is outside $C$.</p>\n      <p>$I=$ [[0]].</p>", "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"answer": "{2*(a1*c1+b1)}*pi*i/{c1}", "showCorrectAnswer": true, "vsetrange": [0, 1], "checkingaccuracy": 0.001, "checkvariablenames": false, "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "type": "jme", "answersimplification": "all", "marks": 1, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "<p>$z=0$ is outside $C$, but $z=\\simplify{1/{c1}}$ is inside $C$.</p>\n      <p>$I=$ [[0]].</p>", "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"answer": "{2*b1}*pi*i/{c1}", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "checkvariablenames": false, "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "type": "jme", "showCorrectAnswer": true, "marks": 1, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "<p>$z=0$ and $z=\\simplify{1/{c1}}$ are both inside $C$.</p>\n      <p>$I=$ [[0]].</p>", "showCorrectAnswer": true, "marks": 0}], "statement": "<p>Evaluate the integral</p>\n    <p>\\[I=\\oint_C{\\!\\simplify{({a1}+{b1}*z)/({c1}*z^2-z)}\\,\\mathrm{d}z},\\]</p>\n    <p>where $C$, mapped counter-clockwise, has the following properties.</p>", "tags": ["checked2015", "MAS210"], "rulesets": {}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "<p><strong>15/7/2012:</strong></p>\n<p>Added tags.</p>", "licence": "Creative Commons Attribution 4.0 International", "description": "<p>Contour integral of a complex-valued function $f(z)$ with the poles of $f(z)$ either inside or outside the path $C$.</p>"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "<h4>Cauchy's Integral Theorem</h4>\n    <blockquote>If $f(z)$ is analytic in a simply connected domain $D$, then for every simple closed path $C$ in $D$ \\[\\oint_C{\\!f(z)\\,\\mathrm{d}z}=0.\\]</blockquote>\n    <h4>Cauchy's Integral Formula</h4>\n    <blockquote>Let $g(z)$ be analytic in a simply connected domain $D$. Then for any point $z_0$ in $D$, and any simple closed path $C$ in $D$ that encloses $z_0$ \\[\\oint_C{\\!\\frac{g(z)}{z-z_0}\\,\\mathrm{d}z}=2\\pi i g(z_0),\\] where the integration is performed counter-clockwise.</blockquote>\n    <h4>Residue Theorem</h4>\n    <blockquote>Let $f(z)$ be analytic inside a closed path $C$, and on $C$, except at finitely many singular points $z_n=z_1,z_2,\\ldots,z_k$ inside $C$. &nbsp;Then the integral of $f(z)$ taken counter-clockwise around $C$ equals $2\\pi i$ times the sum of the residues of $f(z)$ at $z_n=z_1,z_2,\\ldots,z_k$, i.e. \\[\\oint_C{\\!f(z)\\,\\mathrm{d}z}=2\\pi i\\sum_{n=1}^k{\\underset{z=z_n}{\\operatorname{Res}}f(z)}.\\]</blockquote>\n    <p>a)</p>\n    <p>The given function is not analytic at the points $z=0$ and $z=\\simplify{1/{c1}}$.</p>\n    <p>In this part, both of these poles are outside the path $C$. &nbsp;We can therefore use Cauchy's Integral Theorem to state that $I=0$.</p>\n    <p>&nbsp;</p>\n    <p>b)</p>\n    <p>In this part $z=0$ is inside the path $C$, and $z=\\simplify{1/{c1}}$ is outside, so we cannot use the theorem. &nbsp;We can use Cauchy's Integral Formula, however.</p>\n    <p>First rewrite $f(z)$ as $\\frac{g(z)}{z-z_0}$, where $z_0=0$ and $g(z)=\\simplify{({a1}+{b1}*z)/({c1}*z-1)}$.</p>\n    <p>Then</p>\n    <p>\\[I=2\\pi i g(z_0)=\\simplify{{-2*a1}*pi*i}.\\]</p>\n    <p>&nbsp;</p>\n    <p>c)</p>\n    <p>In this part $z=0$ is outside the path $C$, and $z=\\simplify{1/{c1}}$ is inside, so we again cannot use the theorem. &nbsp;We can use Cauchy's Integral Formula, however.</p>\n    <p>First rewrite $f(z)$ as $\\frac{g(z)}{z-z_0}$, where $z_0=\\simplify{1/{c1}}$ and $g(z)=\\simplify{({a1}+{b1}*z)/({c1}*z)}$.</p>\n    <p>Then</p>\n    <p>\\[I=2\\pi i g(z_0)=\\simplify{{2*(a1*c1+b1)}*pi*i/{c1}}.\\]</p>\n    <p>&nbsp;</p>\n    <p>d)</p>\n    <p>In this part, both $z=0$ and $z=\\simplify{1/{c1}}$ are inside the path $C$, so we must use the Residue Theorem to calculate the integral.</p>\n    <p>The residue corresponding to $z=0$ is given by</p>\n    <p>\\[\\underset{z=0}{\\operatorname{Res}}f(z)=\\lim_{z\\rightarrow 0}\\left(\\simplify{({a1}+{b1}*z)/({c1}*z-1)}\\right)=\\var{-a1},\\]</p>\n    <p>and the residue corresponding to $z=\\simplify{1/{c1}}$ is given by</p>\n    <p>\\[\\underset{z=\\simplify{1/{c1}}}{\\operatorname{Res}}f(z)=\\lim_{z\\rightarrow\\simplify{1/{c1}}}\\left(\\simplify{({a1}+{b1}*z)/({c1}*z)}\\right)=\\simplify{{a1*c1+b1}/{c1}}.\\]</p>\n    <p>Therefore</p>\n    <p>\\[I=2\\pi i\\left(\\var{-a1}+\\simplify{{a1*c1+b1}/{c1}}\\right)=\\simplify{{2*b1}*pi*i/{c1}}.\\]</p>", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}