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$\\displaystyle \\simplify[std]{{c1}/{z1}} = $ [[0]]

$\\displaystyle \\simplify[std]{{c2}/{z2}}\\;=\\;$[[0]]

$\\displaystyle \\simplify[std]{{z1}/{z3}}\\;=\\;$[[0]].

Do not include brackets in your answer.

$\\displaystyle \\simplify[std]{{z3}/{z2}}\\;=\\;$[[0]].

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Express the following in the form $a+bi$.

Input $a$ and $b$ as fractions or integers and not as decimals.

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Inverse and division of complex numbers. Four parts.

"}, "functions": {}, "advice": "\n \n \n

Division 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*} \\]

\n \n ", "tags": ["checked2015", "complex numbers", "conjugate of a complex number", "division of complex numbers", "inverse of complex numbers", "multiplication of complex numbers"], "type": "question", "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": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}]}], "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": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}