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<quiz>
  <question type="stack">
    <name>
      <text>10.3.4.2 Hyperbolic function of a complex number</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<p class="HELM_exercise">Exercise</p>
<p>Express <span class="math inline">\(z={@exp@}(a+ib)\)</span> in Cartesian form:</p>
<p>\(z=\quad\)[[input:ans1]][[validation:ans1]] \(\quad+\quad\) [[input:ans2]][[validation:ans2]]\(\ \ i\)&nbsp; &nbsp; [[feedback:prt1]]&nbsp; [[feedback:prt2]]&nbsp;</p><p><strong>Hint:</strong> the imaginary unit, \(i\), should not appear in your final expressions.</p>
<p> Therefore find the real and imaginary parts of \(z = {@exp@}(1 + {@im@}i)\). Express your answers in decimal form, rounding to 3 decimal places if necessary:</p>
<ul>
<li> \(\text{Re}(z)=\quad\)[[input:ans3]][[validation:ans3]][[feedback:prt3]]</li>
<li> \(\text{Im}(z)=\quad\)[[input:ans4]][[validation:ans4]][[feedback:prt4]]</li>
</ul>]]></text>
    </questiontext>
    <generalfeedback format="html">
      <text><![CDATA[<p>We use the hyperbolic sum-angle identity, followed by Osborne's rule:\[{@expr[1]@}(a+ib) \equiv {@expr[1]@}(a)\cosh(ib) +{@expr[6]@}(a)\sinh(ib) \equiv {@ta1@} + i\,{@ta2@}.\]</p>
<p>We then plug in \(a=1\) and \(b={@im@}\) to get:\[{@expr[1]@}(1+{@im@}i) = {@ta3@} + {@ta4@}i.\]</p>]]></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <idnumber/>
    <stackversion>
      <text>2024072400</text>
    </stackversion>
    <questionvariables>
      <text>den: rand([2,3,4,5,6]);
im: %pi/den;
/*Two lists, each corresponding to identity for sinh or cosh*/
expr: rand([[sinh,sinh(a),cos(b),cosh(a),sin(b),cosh,sinh(1),cos(im),cosh(1),sin(im)],[cosh,cosh(a),cos(b),sinh(a),sin(b),sinh,cosh(1),cos(im),sinh(1),sin(im)]]);
/*exp is for the question text*/
exp: expr[1];
ta1: expr[2]*expr[3];
ta2: expr[4]*expr[5];
ta3: ev(0.001*round(1000*expr[7]*expr[8]),simp);
ta4: ev(0.001*round(1000*expr[9]*expr[10]),simp);</text>
    </questionvariables>
    <specificfeedback format="html">
      <text/>
    </specificfeedback>
    <questionnote format="moodle_auto_format">
      <text>\( {@expr[1]@}(a+ib) = {@ta1@} + i{@ta2@} \). For {@im@}: \( \text{Im}(z) = {@ta3@} \), \( \text{Re}(z) = {@ta4@} \).</text>
    </questionnote>
    <questiondescription format="moodle_auto_format">
      <text/>
    </questiondescription>
    <questionsimplify>0</questionsimplify>
    <assumepositive>0</assumepositive>
    <assumereal>0</assumereal>
    <prtcorrect format="html">
      <text><![CDATA[<span style="font-size: 1.5em; color: green; --darkreader-inline-color:#7fff7f;" data-darkreader-inline-color=""><i class="fa fa-check"></i></span> Correct answer, well done.]]></text>
    </prtcorrect>
    <prtpartiallycorrect format="html">
      <text><![CDATA[<span style="font-size: 1.5em; color: orange; --darkreader-inline-color:#ffb733;" data-darkreader-inline-color=""><i class="fa fa-adjust"></i></span> Your answer is partially correct.]]></text>
    </prtpartiallycorrect>
    <prtincorrect format="html">
      <text><![CDATA[<span style="font-size: 1.5em; color: red; --darkreader-inline-color:#ff3333;" data-darkreader-inline-color=""><i class="fa fa-times"></i></span> Incorrect answer.]]></text>
    </prtincorrect>
    <decimals>.</decimals>
    <scientificnotation>*10</scientificnotation>
    <multiplicationsign>none</multiplicationsign>
    <sqrtsign>1</sqrtsign>
    <complexno>i</complexno>
    <inversetrig>cos-1</inversetrig>
    <logicsymbol>lang</logicsymbol>
    <matrixparens>[</matrixparens>
    <isbroken>0</isbroken>
    <variantsselectionseed/>
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      <name>ans1</name>
      <type>algebraic</type>
      <tans>ta1</tans>
      <boxsize>10</boxsize>
      <strictsyntax>1</strictsyntax>
      <insertstars>0</insertstars>
      <syntaxhint/>
      <syntaxattribute>1</syntaxattribute>
      <forbidwords>i,j</forbidwords>
      <allowwords/>
      <forbidfloat>1</forbidfloat>
      <requirelowestterms>0</requirelowestterms>
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      <boxsize>5</boxsize>
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      <forbidwords>sin, cos, sinh, cosh,i,j</forbidwords>
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