// Numbas version: finer_feedback_settings {"name": "Complex Numbers Ex Sheet 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "statement": "

Express the following in the form $a+bi\\;$ where $a$ and $b$ are real.

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Do not include decimals in your answers, only fractions or integers. Also do not include brackets in your answers.

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Elementary examples of multiplication and addition of complex numbers. Four parts.

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Do not include decimals in your answers, only fractions or integers. Also do not include brackets in your answers.

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$(\\simplify[std]{{a}})(\\simplify[std]{{b}})\\;=\\;$[[0]].

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$(\\simplify[std]{{a1}})^2\\;=\\;$[[0]].

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$\\simplify[std,!otherNumbers]{{a3} + {b3} * i + {c3} * i ^ 2 + {d3} * i ^ 3}\\;=\\;$[[0]].

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$(\\simplify[std]{{z1}}) (\\simplify[std]{{z2}}) (\\simplify[std]{{z3}})\\;=\\;$[[0]].

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a)
The formula for multiplying complex numbers is
\\[\\begin{eqnarray*}\\simplify[]{Re((a + ib)(c + id))} &=& ac -bd \\\\ \\simplify[]{Im((a + ib)(c + id))} &=& ad +bc \\end{eqnarray*} \\]

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So we have:
\\[\\begin{eqnarray*}\\simplify[]{Re({a}*{b})} &=& \\simplify[]{{Re(a)}*{Re(b)} - {Im( a)}*{Im(b)} = {Re(a*b)}}\\\\ \\simplify[]{Im({a}*{b})} &=& \\simplify[]{{Re(a)}*{Im(b)} + {Im( a)}*{Re(b)} = {Im(a*b)}} \\end{eqnarray*} \\]
Hence the solution is :

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\\[(\\simplify[std]{{a}})(\\simplify[std]{{b}})=\\var{a*b}\\]
b)

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This is calculated in a similar way once the expression is written as:

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$(\\simplify[std]{{a1}})^2= (\\simplify[std]{{a1}}) (\\simplify[std]{{a1}})$ then we find:

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\\[\\begin{eqnarray*}(\\simplify[std]{{a1}})^2&=& (\\simplify[std]{{a1}}) (\\simplify[std]{{a1}})\\\\ &=& \\simplify[]{({Re(a1)}*{Re(a1)} - {Im(a1)}*{Im(a1)})+ ({Re(a1)}*{Im(a1)} + {Im(a1)}*{Re(a1)})i}\\\\ &=& \\simplify[std]{{a1^2}} \\end{eqnarray*} \\]
c)
We know that $i^2=-1$ which gives $i^3=i^2i=-i$.

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Hence:
\\[ \\begin{eqnarray*} \\simplify[std,!otherNumbers]{{a3} + {b3} * i + {c3} * i ^ 2 + {d3} * i ^ 3}&=&\\simplify[std]{{a3} + {b3} * i -{c3} -({d3} * i)}\\\\ &=&\\simplify[std]{ {a3} -{c3} + ({b3} -{d3}) * i}\\\\ &=&\\simplify[std]{{a3 -c3} + {b3 -d3} * i} \\end{eqnarray*} \\]
d)
This can be calculated by using the formula twice, firstly to multiply out the first two sets of parentheses,
and then to multiply the result of that calculation by the third set of parentheses.

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So we obtain:
\\[ \\begin{eqnarray*} (\\var{z1})(\\var{z2})(\\var{z3})&=&((\\var{z1})(\\var{z2}))(\\var{z3})\\\\ &=&(\\var{z1*z2})(\\var{z3})\\\\ &=&\\var{z1*z2*z3} \\end{eqnarray*} \\]

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