// Numbas version: finer_feedback_settings {"name": "Simon's copy of Polar form of a complex number", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"tags": [], "name": "Simon's copy of Polar form of a complex number", "extensions": [], "variable_groups": [], "rulesets": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "parts": [{"extendBaseMarkingAlgorithm": true, "sortAnswers": false, "variableReplacements": [], "prompt": "

$r=$ [[0]] (Enter your answer to 3 d.p.)

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$\\theta=$ [[0]] (Enter your answer to 3 d.p.)

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Polar form of a complex number.

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Write the complex number $z=\\var{z}$ in polar form $z=r\\mathrm{e}^{i\\theta}$, with $r>0$, $-\\pi<\\theta\\leqslant\\pi$, by calculating $r$ and $\\theta$.

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To write a complex number $z=a+bi$ in polar form $z=r\\mathrm{e}^{i\\theta}$, we calculate the modulus $r = \\lvert z \\rvert$ and argument $\\theta = \\arg(z)$.

Hence

\\[r=\\lvert z \\rvert=\\sqrt{a^2+b^2}=\\sqrt{(\\var{a})^2+(\\var{b})^2}=\\var{absz}\\;\\text{to 3d.p.}\\]

and, in general,

\\[\\theta=\\arg(z)=\\arctan\\left(\\frac{b}{a}\\right).\\]

We have to be careful using this formula for 2 reasons:

Using the arctan function in your calculator will always give $-\\pi/2<\\theta < \\pi/2$, which is not necessarily in the correct quadrant of the Argand diagram. We may have to add or subtract $\\pi$ from $\\arctan\\frac{b}{a}$ in order to obtain $\\arg(z)$ in the correct quadrant (this works because tan is periodic, with period $\\pi$)

In the special case $a=0$, then $\\mathrm{Re}(z)=0$, so $\\arg(z)=\\pm\\frac{\\pi}{2}$, depending on whether $\\mathrm{Im}(z)$ is positive or negative

In our example $a=\\var{a}$, and $b=\\var{b}$, so $\\arctan\\left(\\frac{b}{a}\\right) = \\var{arctan(b/a)}$

Since $a \\var{switch(a<0,\" < \",\" ≥ \")}$ 0 and $b \\var{switch(b<0,\" < \",\" ≥ \")}$ 0, we require $\\var{switch(a<0 and b<0,\" -π < θ < -π/2 \",a<0,\" π/2 < θ ≤ π \",b<0,\" -π/2 ≤ θ < 0 \",\" 0 ≤ θ ≤ π/2 \")}$ to be in the correct quadrant. Hence:

$\\arg(z)=\\var{switch(a<0 and b<0,arctan(b/a),a<0,arctan(b/a),\"(we already have the correct quadrant) = \")} \\var{switch(a<0 and b<0,\" - π =\",a<0,\" + π =\",\"\")}\\var{argz}$.

", "type": "question", "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}]}], "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}