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Find the following complex numbers in the form $a+bi\\;$ where $a$ and $b$ are real.
\nInput all numbers as fractions or integers. Also do not include brackets in your answers.
", "metadata": {"notes": "15/07/2015:
\nAdded tags.
\n4/7/2012
\nAdded tags
\nSame problem in Complex Numbers_2. Question a - sometimes the complex number is generated as a/(b+i*c) but sometimes the complex number is displayed as a decimal, i.e. 0.0975609756+0.1219512195i if this happens then the question is invalid. This is an issue on github.
16/07/2012:
Issue above resolved. Also forbid decimals and brackets. Questions cannot now be answered by simply repeating the expression.
\nAdded formal cba name mas104220122013CBA1_3 as a tag
", "description": "Multiplication and addition of complex numbers. Four parts.
", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "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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