\\[f(z)=\\simplify{{a1}*z^{n1}-{b1}*z^{n1-1}}\\]
\n$I=$ [[0]].
", "showCorrectAnswer": true, "marks": 0}, {"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": "\\[f(z)=\\simplify{{a2}*z^{n2}/(z-{b2})}\\]
\n$I=$ [[0]].
", "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"answer": "i*pi*{2*a3}/({b3}^{1+n3})", "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": "\\[f(z)=\\simplify{{a3}*z^{n3}/({b3}*z-1)}\\]
\n$I=$ [[0]].
", "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"answer": "2*pi/{a4^(n4+1)}*(({in4p1*a4switch[1]}/2+{in4p1*a4switch[2]}/sqrt(2))+({in4p2*a4switch[0]}+{in4p2*a4switch[1]}*sqrt(3)/2+{in4p2*a4switch[2]}/sqrt(2)))", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "checkvariablenames": false, "expectedvariablenames": [], "notallowed": {"message": "Do not enter decimals, the exponential function, or any trigonometric functions in your answer.
", "showStrings": false, "partialCredit": 0, "strings": [".", "e", "exp", "cos", "sin", "tan", "sec", "cosec", "cot"]}, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "type": "jme", "showCorrectAnswer": true, "marks": 1, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\\[f(z)=\\frac{\\simplify{z^{n4}}\\mathrm{e}^{\\pi z}}{\\simplify{{a4}*z}-i}\\]
\n$I=$ [[0]].
\nDo not use decimals, the exponential function, or any trigonometric functions in your answer. If you need to enter $\\pi$ write pi, and if you need to enter a square root, e.g. $\\sqrt{x}$, enter sqrt(x).
For each of the following functions $f(z)$, use Cauchy's Integral Theorem or Cauchy's Integral Formula (as appropriate) to evaluate
\n\\[I=\\oint_C{\\!f(z)\\,\\mathrm{d}z},\\]
\nwhere $C$ is the unit circle $\\lvert z\\rvert=1$ mapped counter-clockwise.
", "tags": ["MAS2103", "checked2015"], "rulesets": {}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "15/7/2012:
\nAdded tags.
", "licence": "Creative Commons Attribution 4.0 International", "description": "Cauchy's integral theorem/formula for several functions $f(z)$ and $C$ the unit circle.
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "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.\\]\n
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.\n
a)
\n$f(z)=\\simplify{{a1}*z^{n1}-{b1}*z^{n1-1}}$ is analytic for all $z$, so we can use the theorem to obtain $I=0$.
\n\n
b)
\n$f(z)=\\simplify{{a2}*z^{n2}/(z-{b2})}$ is not analytic at $z=\\var{b2}$, but this point is outside $C$, so we can again use the theorem to obtain $I=0$.
\n\n
c)
\n$f(z)=\\simplify{{a3}*z^{n3}/({b3}*z-1)}$ is not analytic at $z=\\simplify{1/{b3}}$, and this point is inside $C$, so we cannot use the theorem.
\nInstead, we must use the formula, so
\n\\[I=\\oint{\\!\\frac{g(z)}{z-z_0}\\,\\mathrm{d}z}=2\\pi i g(z_0),\\]
\nwhere $g(z)=\\simplify{{a3}/{b3}*z^{n3}}$ and $z_0=\\simplify{1/{b3}}$, and therefore
\n\\[I=2\\pi i\\times\\simplify{{a3}/{b3}}\\times\\simplify{1/{b3}^{n3}}=\\simplify{i*pi*{2*a3}/({b3}^{1+n3})}.\\]
\n\n
d)
\n$f(z)=\\frac{\\simplify{z^{n4}}\\mathrm{e}^{\\pi z}}{\\simplify{{a4}*z}-i}$ is not anayltic at $z=\\frac{i}{\\var{a4}}$, and this point is inside $C$, so we again cannot use the theorem.
\nUsing the formula, we have $g(z)=\\simplify{1/{a4}z^{n4}}\\mathrm{e}^{\\pi z}$, and $z_0=\\frac{i}{\\var{a4}}$, so
\n\\[I=2\\pi i\\times\\simplify{1/{a4}}\\times\\simplify{(i/{a4})^{n4}}\\mathrm{e}^{\\frac{i\\pi}{\\var{a4}}}=\\simplify{2*pi/{a4^(n4+1)}*(({in4p1*a4switch[1]}/2+{in4p1*a4switch[2]}/sqrt(2))+({in4p2*a4switch[0]}+{in4p2*a4switch[1]}*sqrt(3)/2+{in4p2*a4switch[2]}/sqrt(2)))}.\\]
\n ", "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/"}]}