# Test: Complex Numbers IV - Ambitious

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Question 1:   Find $3$ cube roots of $z=8\left(\mathrm{cos}81°+i\mathrm{sin}81°\right)$
$\left\{\begin{array}{l}{z}_{1}=8\overline{)27°}\\ {z}_{2}=8\overline{)147°}\\ {z}_{3}=8\overline{)267°}\end{array}$
$\left\{\begin{array}{l}{z}_{1}=2\overline{)27°}\\ {z}_{2}=2\overline{)147°}\\ {z}_{3}=2\overline{)267°}\end{array}$
$\left\{\begin{array}{l}{z}_{1}=2\overline{)27°}\\ {z}_{2}=2\overline{)54°}\\ {z}_{3}=2\overline{)81°}\end{array}$
$\left\{\begin{array}{l}{z}_{1}=8\overline{)27°}\\ {z}_{2}=8\overline{)54°}\\ {z}_{3}=8\overline{)81°}\end{array}$
Question 2:   Find $5$ fifth roots of $z=14\overline{)305°}$
$\left\{\begin{array}{l}z{}_{1}=2\overline{)61°}\\ z{}_{2}=2\overline{)133°}\\ z{}_{3}=2\overline{)203°}\\ z{}_{4}=2\overline{)274°}\\ z{}_{5}=2\overline{)345°}\end{array}$
$\left\{\begin{array}{l}z{}_{1}=1.695\overline{)60°}\\ z{}_{2}=1.695\overline{)120°}\\ z{}_{3}=1.695\overline{)180°}\\ z{}_{4}=1.695\overline{)240°}\\ z{}_{5}=1.695\overline{)300°}\end{array}$
$\left\{\begin{array}{l}z{}_{1}=1.695\overline{)61°}\\ z{}_{2}=1.695\overline{)133°}\\ z{}_{3}=1.695\overline{)205°}\\ z{}_{4}=1.695\overline{)277°}\\ z{}_{5}=1.695\overline{)349°}\end{array}$
$\left\{\begin{array}{l}z{}_{1}=2\overline{)61°}\\ z{}_{2}=2\overline{)111°}\\ z{}_{3}=2\overline{)161°}\\ z{}_{4}=2\overline{)201°}\\ z{}_{5}=2\overline{)251°}\end{array}$
Question 3:   Find the principal root of the $3$ cube roots of the $z=5\left(\mathrm{cos}270°+i\mathrm{sin}270°\right)$
$1.71\overline{)210°}$
$1.71\overline{)90°}$
$1.71\overline{)330°}$
$\frac{5}{3}\overline{)90°}$
Question 4:   Find the expression for $\mathrm{sin}4\theta$ in terms of $\mathrm{sin}\theta$ (De Moivre’s theorem)
$3{\mathrm{cos}}^{4}\theta +8{\mathrm{cos}}^{3}\theta +1$
$8{\mathrm{cos}}^{4}\theta -8{\mathrm{cos}}^{3}\theta +2$
$8{\mathrm{cos}}^{4}\theta -8{\mathrm{cos}}^{2}\theta +1$
$2\mathrm{cos}\theta -3\mathrm{cos}\theta +5$
Question 5:   Find the expression for ${\mathrm{sin}}^{3}\theta$
${\mathrm{sin}}^{3}\theta =\frac{1}{7}i\mathrm{sin}5\theta +\frac{1}{6}i\mathrm{sin}3\theta$
${\mathrm{sin}}^{3}\theta =\frac{1}{4}i\mathrm{sin}3\theta -\frac{3}{4}i\mathrm{sin}\theta$
${\mathrm{sin}}^{3}\theta =-2i\mathrm{sin}3\theta -\frac{1}{2}i\mathrm{sin}\theta$
${\mathrm{sin}}^{3}\theta =-\frac{2}{5}i\mathrm{sin}3\theta -\frac{1}{4}i\mathrm{sin}2\theta$
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