91Ó°ÊÓ

Recall from the introduction to Section \(8.2\) that Jerome Cardan's solutions to the equation \(x^{3}=15 x+4\) could be written as $$ x=\sqrt[3]{2+11 i}+\sqrt[3]{2-11 i} $$ Let's assume that the cube roots shown above arc complex conjugates. If they are, then we can simplify our work by noticing that $$ x=\sqrt[3]{2+11 i}+\sqrt[3]{2-11 i}=a+b i+a-b i=2 a $$ which means that we simply double the real part of each cube root of \(2+11 i\) to find the solutions to \(x^{3}=15 x+4\). Now, to end our work with Cardan, find the 3 cube roots of \(2+11 i\). Then, noting the discussion above, use the 3 cube roots to solve the equation \(x^{3}=15 x+4\). Write your answers accurate to the nearest thousandth.

Find the quotient \(z_{1} / z_{2}\) in standard form. Then write \(z_{1}\) and \(z_{2}\) in trigonometric form and find their quotient again. Finally, convert the answer that is in trigonometric form to standard form to show that the two quotients are equal. $$ z_{1}=8, z_{2}=-4 $$

$$ \text { Change each equation to rectangular coordinates and then graph. } $$ $$ r(1-\cos \theta)=1 $$

Find the following products. \((3+2 i)^{2}\)

Write each complex number in trigonometric form. Round all angles to the nearest hundredth of a degree.\(3+4 i\)

Write each complex number in trigonometric form. Round all angles to the nearest hundredth of a degree.\(3-4 i\)

Find the following products. \((3-2 i)^{2}\)

$$ \text { Change each equation to rectangular coordinates and then graph. } $$ $$ r(1-\sin \theta)=1 $$

Find the quotient \(z_{1} / z_{2}\) in standard form. Then write \(z_{1}\) and \(z_{2}\) in trigonometric form and find their quotient again. Finally, convert the answer that is in trigonometric form to standard form to show that the two quotients are equal. $$ z_{1}=-6, z_{2}=3 $$

\(r^{2}=9\)