// Numbas version: finer_feedback_settings {"name": "Paul's copy of Arithmetics of complex numbers I", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Paul's copy of Arithmetics of complex numbers I", "tags": ["addition of complex numbers", "checked2015", "complex numbers", "multiplication of complex numbers", "product of complex numbers"], "metadata": {"description": "
Elementary examples of multiplication and addition of complex numbers. Four parts.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Express the following in the form $x+iy\\;$ where $x$ and $y$ are real numbers.
\nWrite your answers using only integers or fractions, not decimals. To input, e.g., $2+3i$, type 2+3i putting the i last.
\nTo input, e.g., $\\frac{3}{10}+\\frac{4}{5}i$, type 3/10+4/5i or (3/10)+(4/5)i. There should be no brackets in your answer except to indicate fractions as here.
", "advice": "a)
\nMultiplying two complex numbers and extracting their real and imaginary parts, we find:
\n\\[\\begin{eqnarray*}\\simplify[]{Re((x_1 + iy_1)(x_2 + iy_2))} &=& x_1 x_2 -y_1y_2 \\\\ \\simplify[]{Im((x_1 + iy_1)(x_2 + iy_2))} &=& x_1 y_2 +y_1x_2 \\end{eqnarray*} \\]
\nSo 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 :
\\[(\\simplify[std]{{a}})(\\simplify[std]{{b}})=\\var{a*b}\\]
b)
This is calculated in a similar way once the expression is written as:
\n$(\\simplify[std]{{a1}})^2= (\\simplify[std]{{a1}}) (\\simplify[std]{{a1}})$ then we find:
\n\\[\\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 $i^3=i^2i=-i$.
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.
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*} \\]
$(\\simplify[std]{{a}})(\\simplify[std]{{b}})\\;=\\;$[[0]].
\n\n
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