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Firstly, $|z| = \\sqrt{(\\var{a1})^2+(\\var{b1})^2} = \\sqrt{\\var{a1*a1+b1*b1}} = \\var{dpformat(modz,2)}$
\nSecondly, since the real part of $z$ is {rez} and the imaginary part of $z$ is {imz}, {quad} So,
\n$\\arg(z) = \\var{dpformat(argz,2)}^\\circ$.
\n
im
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\n$|z| = $ [[0]]
\n$\\arg(z) = $ [[1]]$^\\circ$.
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", "licence": "Creative Commons Attribution 4.0 International"}, "preamble": {"css": "", "js": ""}, "type": "question", "contributors": [{"name": "Peter Johnston", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/771/"}]}]}], "contributors": [{"name": "Peter Johnston", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/771/"}]}