// Numbas version: finer_feedback_settings {"name": "WM104 in-class assessment: part (i) - complex numbers", "metadata": {"description": "
Content assessed : complex arithmetic; argument and modulus of complex numbers; de Moivre's theorem.
This complex numbers in-class assesment counts 20% towards your final maths grade for WM104.
Note that although questions are randomised for each student, all questions test the same learning outcomes at the same level for each student.
If you have any questions during the test, please put up your hand to alert the invigilator that you need attention.
", "licence": "Creative Commons Attribution 4.0 International"}, "duration": 1500, "percentPass": "40", "showQuestionGroupNames": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questions": [{"name": "Shaheen's copy of Arithmetics of complex numbers II", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Shaheen Charlwood", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1819/"}], "tags": [], "metadata": {"description": "Multiplication and addition of complex numbers. Four parts.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Find the following complex numbers in the form $a+bi\\;$ where $a$ and $b$ are real.
\nInput all numbers as fractions or integers (whole numbers). Also do not include brackets in your answers.
", "advice": "a)
\nThe solution is given by:
\n
$\\simplify[std]{{e6*i}}(\\simplify[std]{{a}})=\\simplify[std]{{a*e6*i}}$
b)
$\\simplify[std]{{a}*{z4}={a*z4}}$
\n
c)
\\[ \\begin{eqnarray*} \\simplify[std,!otherNumbers]{{a}*({a3} + {b3} * i + {c3} * i ^ 2 + {d3} * i ^ 3)}&=&\\simplify[std]{{a}*{a3 + b3 * i + c3 * i ^ 2 + d3 * i ^ 3}}\\\\ &=&\\simplify[std]{{a*(a3 + b3 * i + c3 * i ^ 2 + d3 * i ^ 3)}} \\end{eqnarray*} \\]
d)
This can be calculated by using the formula twice, firstly to multiply out the first two sets of parentheses,
\nand then to multiply the result of that calculation by the third set of parentheses.
\nSo we obtain:
\\[ \\begin{eqnarray*} (\\var{a})(\\var{z1})(\\var{z3})&=&((\\var{a})(\\var{z1}))(\\var{z3})\\\\ &=&(\\var{a*(z1)})(\\var{z3})\\\\ &=&\\var{a*(z1)*(z3)} \\end{eqnarray*} \\]
$\\var{e6*i}(\\simplify[std]{{a}})\\;=\\;$[[0]].
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"}, "valuegenerators": []}], "sortAnswers": false}], "type": "question"}, {"name": "Inverse and division of complex numbers", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Shaheen Charlwood", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1819/"}], "tags": [], "metadata": {"description": "Two parts.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Express the following in the form $a+bi$.
\nInput $a$ and $b$ as fractions or integers (whole numbers) and not as decimals.
", "advice": "\n \n \nDivision of two complex numbers can be performed by mutiplying both the numerator and denominator by the conjugate of the denominator.
Suppose that \\[ z = \\frac{a+bi}{c+di},\\;\\; c+di \\neq 0\\] then we have:
\\[\\begin{eqnarray*}\n \n z&=&\\frac{a+bi}{c+di}\\\\\n \n &=&\\frac{(a+bi)(c-di)}{(c+di)(c-di)}\\\\\n \n &=&\\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}\\\\\n \n &=&\\frac{ac+bd}{c^2+d^2}+\\frac{bc-ad}{c^2+d^2}i\n \n \\end{eqnarray*}\n \n \\]
Although this is a formula for the inverse, the best way to find these complex numbers is to remember to multiply top and bottom by the conjugate of the denominator.
(a)
\\[\\begin{eqnarray*}\\simplify[std]{{c1}/{z1}} &=&\\simplify[std]{({c1}*{conj(z1)})/({z1}*{conj(z1)})}\\\\\n \n &=&\\simplify[std]{{c1*conj(z1)}/{abs(z1)^2}}\\\\\n \n &=& \\simplify[std]{{c1*re(z1)}/{abs(z1)^2}-{c1*im(z1)}/{abs(z1)^2}*i}\n \n \\end{eqnarray*} \\]
(b)
\\[\\begin{eqnarray*}\\simplify[std]{{c2}/{z2}} &=&\\simplify[std]{({c2}*{conj(z2)})/({z2}*{conj(z2)})}\\\\\n \n &=&\\simplify[std]{{c2*conj(z2)}/{abs(z2)^2}}\\\\\n \n &=& \\simplify[std]{{c2*re(z2)}/{abs(z2)^2}-{c2*im(z2)}/{abs(z2)^2}*i}\n \n \\end{eqnarray*} \\]
(c)
\\[\\begin{eqnarray*}\\simplify[std]{{z1}/{z3}} &=&\\simplify[std]{({z1}*{conj(z3)})/({z3}*{conj(z3)})}\\\\\n \n &=&\\simplify[std]{{z1*conj(z3)}/{abs(z3)^2}}\\\\\n \n &=& \\simplify[std]{{re(z1*conj(z3))}/{abs(z3)^2}+{im(z1*conj(z3))}/{abs(z3)^2}*i}\n \n \\end{eqnarray*} \\]
(d)
\\[\\begin{eqnarray*}\\simplify[std]{{z3}/{z2}} &=&\\simplify[std]{({z3}*{conj(z2)})/({z2}*{conj(z2)})}\\\\\n \n &=&\\simplify[std]{{z3*conj(z2)}/{abs(z2)^2}}\\\\\n \n &=& \\simplify[std]{{re(z3*conj(z2))}/{abs(z2)^2}+{im(z3*conj(z2))}/{abs(z2)^2}*i}\n \n \\end{eqnarray*} \\]
$\\displaystyle \\simplify[std]{{c1}/{z1}} = $ [[0]]
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\nDo not include brackets in your answer.
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\n$z=\\;\\;$[[0]].
\n ", "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "useCustomName": false, "customName": "", "unitTests": [], "showFeedbackIcon": true, "scripts": {}, "gaps": [{"answer": "{re(conj(z1)*z3*z2)}/{abs(z1)^2}+{im(conj(z1)*z3*z2)}/{abs(z1)^2}*i", "mustmatchpattern": {"message": "Your answer is not in the form $a+bi$.", "pattern": "`+-((integer:$n/integer:$n`?))`? + ((`+-integer:$n`?/integer:$n`?)*i `| `+-i)`?", "partialCredit": 0, "nameToCompare": ""}, "vsetRangePoints": 5, "useCustomName": false, "checkingType": "absdiff", "valuegenerators": [], "vsetRange": [0, 1], "showFeedbackIcon": true, "type": "jme", "variableReplacementStrategy": "originalfirst", "checkingAccuracy": 0.001, "variableReplacements": [], "failureRate": 1, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "showPreview": true, "customName": "", "checkVariableNames": false, "unitTests": [], "scripts": {}, "answerSimplification": "std", "showCorrectAnswer": true, "marks": 1}], "type": "gapfill", "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "variableReplacements": [], "marks": 0, "sortAnswers": false}, {"prompt": "\n\\[\\displaystyle z=\\simplify[!collectNumbers]{({z2}*{z1})}(\\var{z3})^{-1}\\]
\n$z=\\;\\;$[[0]].
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\nInput $a$ and $b$ as fractions and not as decimals.
", "tags": ["algebra of complex numbers", "checked2015", "complex arithmetic", "complex numbers", "division of complex numbers", "inverse of complex numbers", "multiplication of complex numbers", "product of complex numbers"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus", "!collectlikefractions"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Composite multiplication and division of complex numbers. Two parts.
"}, "advice": "\na)
\\[\\begin{eqnarray*}z=\\simplify[!collectNumbers]{({z3}*{z2})/{z1}} &=&\\simplify[!collectNumbers]{({z3}*{z2}*{conj(z1)})/({z1}*{conj(z1)})}\\\\ &=&\\simplify[!collectNumbers]{({z3*z2}*{conj(z1)})/({abs(z1)^2})}\\\\ &=&\\simplify[!collectNumbers]{{z3*z2*conj(z1)}/{abs(z1)^2}}\\\\ &=& \\simplify[std]{{re(z3*z2*conj(z1))}/{abs(z1)^2}+{im(z3*z2*conj(z1))}/{abs(z1)^2}*i} \\end{eqnarray*} \\]
b)
\\[\\begin{eqnarray*}z= \\simplify[!collectNumbers]{({z2}*{z1})}(\\var{z3})^{-1} &=& \\simplify[!collectNumbers]{({z2}*{z1})/{z3}}\\\\ &=&\\simplify[!collectNumbers]{({z2}*{z1}*{conj(z3)})/({z3}*{conj(z3)})}\\\\ &=&\\simplify[!collectNumbers]{({z2*z1}*{conj(z3)})/({abs(z3)^2})}\\\\ &=&\\simplify[!collectNumbers]{{z2*z1*conj(z3)}/{abs(z3)^2}}\\\\ &=& \\simplify[std]{{re(z2*z1*conj(z3))}/{abs(z3)^2}+{im(z2*z1*conj(z3))}/{abs(z3)^2}*i} \\end{eqnarray*} \\]
$|\\var{z1}|=\\;\\;$[[0]], $\\arg(\\var{z1})=\\;\\;$[[1]] radians
\nInput both answers to 3 decimal places.
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\nInput both answers to 3 decimal places.
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\nInput both answers to 3 decimal places.
", "scripts": {}, "gaps": [{"correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "allowFractions": false, "type": "numberentry", "maxValue": "ans4+tol", "minValue": "ans4-tol", "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "marks": 1, "showPrecisionHint": false}, {"correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "allowFractions": false, "type": "numberentry", "maxValue": "arg4+tol", "minValue": "arg4-tol", "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "marks": 1, "showPrecisionHint": false}], "type": "gapfill", "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "marks": 0}], "statement": "Find the modulus and argument (in radians) of the following complex numbers, where the argument lies between $-\\pi$ and $\\pi$.
\nWhen calculating the argument pay particular attention to the quadrant in which the complex number lies.
\nInput all answers to 3 decimal places.
", "tags": ["arctan", "arg", "argument", "argument of complex numbers", "checked2015", "complex number", "complex numbers", "mas1602", "MAS1602", "mod", "modulus", "modulus argument form", "modulus of complex numbers", "quadrants and complex numbers"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "15/7/2015:
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\n5/07/2012:
\nAdded tags.
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\nQuestion appears to be working correctly.
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", "licence": "Creative Commons Attribution 4.0 International", "description": "Finding the modulus and argument (in radians) of four complex numbers; the arguments between $-\\pi$ and $\\pi$ and careful with quadrants!
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "Note that the arguments $\\theta$ of the complex numbers are in radians and have to be in the range $-\\pi < \\theta \\le \\pi$.
\nYou have to be careful with using a standard calculator when you are finding the argument of a complex number.
\nIf $z=a+bi=r(\\cos(\\theta)+i\\sin(\\theta))$ then we have:$r\\cos(\\theta)=a,\\;\\;r\\sin(\\theta)=b$ and so $\\tan(\\theta) = b/a$.
\nUsing a calculator to find the argument via $\\arctan(b/a)$ works in the range $-\\pi < \\theta \\le \\pi$ when the complex number is in the first or fourth quadrants – you get the correct value.
\nHowever, The calculator gives the wrong value for complex numbers in the other quadrants.
\nComplex number in the Second Quadrant.
\nSince $\\arctan(b/a)$ does not distinguish between the second and fourth quadrants and the calculator gives the argument for the fourth quadrant you have to add $\\pi$ onto the calculator value.
\nComplex number in the Third Quadrant.
\nSince $\\arctan(b/a)$ does not distinguish between the first and third quadrants and the calculator gives the argument for the first quadrant you have to take away $\\pi$ from the calculator value.
\na)Modulus.
\n\\[ \\begin{eqnarray*} |\\var{z1}|&=&\\sqrt{(\\var{a1})^2+(\\var{b1})^2}\\\\ &=& \\var{abs(z1)}\\\\ &=&\\var{ans1} \\end{eqnarray*} \\] to 3 decimal places.
\nArgument.
\n{m1}
\nHence we see that: \\[\\begin{eqnarray*} \\arg(\\var{z1}) &=& \\var{arg(z1)}\\\\ &=& \\var{arg1}\\; \\mbox{radians} \\end{eqnarray*} \\] to 3 decimal places.
\nb)Modulus.
\n\\[ \\begin{eqnarray*} |\\var{z2}|&=&\\sqrt{(\\var{a2})^2+(\\var{b2})^2}\\\\ &=& \\var{abs(z2)}\\\\ &=&\\var{ans2} \\end{eqnarray*} \\] to 3 decimal places.
\nArgument.
\n{m2}
\nHence we see that: \\[\\begin{eqnarray*} \\arg(\\var{z2}) &=& \\var{arg(z2)}\\\\ &=& \\var{arg2}\\; \\mbox{radians} \\end{eqnarray*} \\] to 3 decimal places.
\nc)Modulus.
\n\\[ \\begin{eqnarray*} |\\var{z3}|&=&\\sqrt{(\\var{c2})^2+(\\var{d2})^2}\\\\ &=& \\var{abs(z3)}\\\\ &=&\\var{ans3} \\end{eqnarray*} \\] to 3 decimal places.
\nArgument.
\n{m3}
\nHence we see that: \\[\\begin{eqnarray*} \\arg(\\var{z3}) &=& \\var{arg(z3)}\\\\ &=& \\var{arg3}\\; \\mbox{radians} \\end{eqnarray*} \\] to 3 decimal places.
\nd)Modulus.
\n\\[ \\begin{eqnarray*} |\\var{z4}|&=&\\sqrt{(\\var{a3})^2+(\\var{b3})^2}\\\\ &=& \\var{abs(z4)}\\\\ &=&\\var{ans4} \\end{eqnarray*} \\] to 3 decimal places.
\nArgument.
\n{m4}
\nHence we see that: \\[\\begin{eqnarray*} \\arg(\\var{z4}) &=& \\var{arg(z4)}\\\\ &=& \\var{arg4}\\; \\mbox{radians} \\end{eqnarray*} \\] to 3 decimal places.
"}, {"name": "Shaheen's copy of De Moivre's Theorem: Positive Powers", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Shaheen Charlwood", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1819/"}], "tags": [], "metadata": {"description": "Find modulus and argument of two complex numbers.
Then use De Moivre's Theorem to find powers of the complex numbers.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Use de Moivre's theorem to write the following complex number in the form $a+bi$.
\nRemember that the argument of a complex numbers lies in the range $-\\pi \\lt \\theta \\le \\pi$.
\nImportant
\nWhen calculating the final answer in part (iii), you must use whole number values (integers), otherwise the final answer will not be marked as correct.
", "advice": "Given a complex number $z=r(\\cos(\\theta)+i\\sin(\\theta))$ de Moivre's theorem states that $z^n=r^n(\\cos(n\\theta)+i\\sin(n\\theta))$ for an integer power $n$.
So if we know the modulus $r$ and the argument $\\theta$ for $z$ then the theorem provides a way of calculating $z^n$.
As usual, you must be careful that the argument is calculated correctly by paying attention to the quadrant of the complex plane in which the complex number lies.
\nAlso remember that for this question, arguments of complex numbers lie in the range $-\\pi \\lt \\theta \\le \\pi$.
\nWith the above in mind we can now answer the questions:
\na)
\n\\[ \\begin{eqnarray*} |\\var{z1}|&=&\\sqrt{(\\var{a1})^2+(\\var{b1})^2}\\\\ &=& \\var{abs(z1)}\\\\ &=&\\var{ans1} \\end{eqnarray*} \\] to 3 decimal places.
\nNote that $r^{\\var{n2}}=|(\\var{z1})^{\\var{n2}}| =\\var{abs(z1)}^{\\var{n2}}=\\var{abs(z1)^n2}$ which we will use in the calculation for $(\\var{z1})^{\\var{n2}}$
\n{m1}.
Hence we see that:
\\[\\begin{eqnarray*} \\arg(\\var{z1}) &=& \\var{arg(z1)}\\\\ &=& \\var{arg1}\\; \\mbox{radians} \\end{eqnarray*} \\] to 3 decimal places.
We have $\\arg((\\var{z1})^{\\var{n2}})=\\var{n2}\\times \\var{arg(z1)} = \\var{n2*arg(z1)}$ radians.
\nHence we have \\[\\begin{eqnarray*}(\\var{z1})^{\\var{n2}} &=& \\var{abs(z1)^n2}(\\cos(\\var{n2*arg(z1)})+\\sin(\\var{n2*arg(z1)})i)\\\\ &=& \\var{abs(z1)^n2}\\cos(\\var{n2*arg(z1)})+\\var{abs(z1)^n2}\\times\\sin(\\var{n2*arg(z1)})i\\\\ &=& \\simplify[std]{{a3}+{b3}i}. \\end{eqnarray*} \\]
\nb)
\n\\[ \\begin{eqnarray*} |\\var{z2}|&=&\\sqrt{(\\var{a2})^2+(\\var{b2})^2}\\\\ &=& \\var{abs(z2)}\\\\ &=&\\var{ans2} \\end{eqnarray*} \\] to 3 decimal places.
\nNote that $r^{\\var{n4}}=|(\\var{z2})^{\\var{n4}}| =\\var{abs(z2)}^{\\var{n4}}=\\var{abs(z2)^n4}$ which we will use in the calculation for $(\\var{z2})^{\\var{n4}}$
\n{m2}.
Hence we see that:
\\[\\begin{eqnarray*} \\arg(\\var{z2}) &=& \\var{arg(z2)}\\\\ &=& \\var{arg2}\\; \\mbox{radians} \\end{eqnarray*} \\] to 3 decimal places.
We have $\\arg((\\var{z2})^{\\var{n4}})=\\var{n4}\\times \\var{arg(z2)} = \\var{n4*arg(z2)}$ radians.
\nHence we have \\[\\begin{eqnarray*}(\\var{z2})^{\\var{n4}} &=& \\var{abs(z2)^n4}(\\cos(\\var{n4*arg(z2)})+\\sin(\\var{n4*arg(z2)})i)\\\\ &=& \\var{abs(z2)^n4}\\cos(\\var{n4*arg(z2)})+\\var{abs(z2)^n4}\\times\\sin(\\var{n4*arg(z2)})i\\\\ &=& \\simplify[std]{{a4}+{b4}i}. \\end{eqnarray*} \\]
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Find the modulus and argument of $\\var{z1}$ correct to 3 decimal places.
\n(i) $|\\var{z1}|\\;=\\;$ [[0]], correct to 3 decimal places.
\n(ii) $\\arg(\\var{z1})\\;=\\;$[[1]] radians, correct to 3 decimal places.
\nHence find:
\n(iii) $(\\var{z1})^{\\var{n2}}\\;=\\;$[[2]]
\nInput as a complex number, with integer values for the real and imaginary parts.
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Find the following complex numbers in the form $a+bi\\;$ where $a$ and $b$ are real.
\nInput all numbers as fractions or integers (whole numbers). Also do not include brackets in your answers.
", "advice": "a)
\nThe solution is given by:
\n
$\\simplify[std]{{e6*i}}(\\simplify[std]{{a}})=\\simplify[std]{{a*e6*i}}$
b)
$\\simplify[std]{{a}*{z4}={a*z4}}$
\n
c)
\\[ \\begin{eqnarray*} \\simplify[std,!otherNumbers]{{a}*({a3} + {b3} * i + {c3} * i ^ 2 + {d3} * i ^ 3)}&=&\\simplify[std]{{a}*{a3 + b3 * i + c3 * i ^ 2 + d3 * i ^ 3}}\\\\ &=&\\simplify[std]{{a*(a3 + b3 * i + c3 * i ^ 2 + d3 * i ^ 3)}} \\end{eqnarray*} \\]
d)
This can be calculated by using the formula twice, firstly to multiply out the first two sets of parentheses,
\nand then to multiply the result of that calculation by the third set of parentheses.
\nSo we obtain:
\\[ \\begin{eqnarray*} (\\var{a})(\\var{z1})(\\var{z3})&=&((\\var{a})(\\var{z1}))(\\var{z3})\\\\ &=&(\\var{a*(z1)})(\\var{z3})\\\\ &=&\\var{a*(z1)*(z3)} \\end{eqnarray*} \\]
$\\var{e6*i}(\\simplify[std]{{a}})\\;=\\;$[[0]].
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Find the following complex numbers in the form $a+bi\\;$ where $a$ and $b$ are real.
\nInput all numbers as fractions or integers (whole numbers). Also do not include brackets in your answers.
", "advice": "a)
\nThe solution is given by:
\n
$\\simplify[std]{{e6*i}}(\\simplify[std]{{a}})=\\simplify[std]{{a*e6*i}}$
b)
$\\simplify[std]{{a}*{z4}={a*z4}}$
\n
c)
\\[ \\begin{eqnarray*} \\simplify[std,!otherNumbers]{{a}*({a3} + {b3} * i + {c3} * i ^ 2 + {d3} * i ^ 3)}&=&\\simplify[std]{{a}*{a3 + b3 * i + c3 * i ^ 2 + d3 * i ^ 3}}\\\\ &=&\\simplify[std]{{a*(a3 + b3 * i + c3 * i ^ 2 + d3 * i ^ 3)}} \\end{eqnarray*} \\]
d)
This can be calculated by using the formula twice, firstly to multiply out the first two sets of parentheses,
\nand then to multiply the result of that calculation by the third set of parentheses.
\nSo we obtain:
\\[ \\begin{eqnarray*} (\\var{a})(\\var{z1})(\\var{z3})&=&((\\var{a})(\\var{z1}))(\\var{z3})\\\\ &=&(\\var{a*(z1)})(\\var{z3})\\\\ &=&\\var{a*(z1)*(z3)} \\end{eqnarray*} \\]
$\\var{e6*i}(\\simplify[std]{{a}})\\;=\\;$[[0]].
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