Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.3

Question 3

auto_storiesAcademic Material

Find roots and express as a product of linear factors:z4+7z2144=0z^4 + 7z^2 - 144 = 0.

Solution

Let u=z2 to form a quadratic equation:z4+7z2144=0    u2+7u144=0Quadratic formula: u=b±b24ac2au=7±494(1)(144)2(1)=7±49+5762=7±6252=7±252u=7+252=182=9oru=7252=322=16Since u=z2, we have z2=9 or z2=16z2=9    z=±3z2=16    z=±4i(since 16=±4i)Therefore, the roots are: 3,3,4i,4iUsing the factor theorem:z4+7z2144=(z3)(z+3)(z4i)(z+4i)\begin{aligned} & \boxed{\text{Let } u = z^2 \text{ to form a quadratic equation:}} \\ \\ & z^4 + 7z^2 - 144 = 0 \implies u^2 + 7u - 144 = 0 \\ \\ & \boxed{\text{Quadratic formula: } u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}} \\ \\ & u = \frac{-7 \pm \sqrt{49 - 4(1)(-144)}}{2(1)} = \frac{-7 \pm \sqrt{49 + 576}}{2} = \frac{-7 \pm \sqrt{625}}{2} = \frac{-7 \pm 25}{2} \\ \\ & u = \frac{-7 + 25}{2} = \frac{18}{2} = 9 \quad \text{or} \quad u = \frac{-7 - 25}{2} = \frac{-32}{2} = -16 \\ \\ & \boxed{\text{Since } u = z^2 \text{, we have } z^2 = 9 \text{ or } z^2 = -16} \\ \\ & z^2 = 9 \implies z = \pm 3 \\ \\ & z^2 = -16 \implies z = \pm 4i \quad (\text{since } \sqrt{-16} = \pm 4i) \\ \\ & \boxed{\text{Therefore, the roots are: } 3, -3, 4i, -4i} \\ \\ & \boxed{\text{Using the factor theorem:}} \\ \\ & z^4 + 7z^2 - 144 = (z - 3)(z + 3)(z - 4i)(z + 4i) \end{aligned}