## Precalculus (6th Edition) Blitzer

Solutions of the provided equation in polar form are: \begin{align} & 2\left[ \cos 67.5{}^\circ +i\sin 67.5{}^\circ \right],2\left[ \cos 157.5{}^\circ +i\sin 157.5{}^\circ \right],2\left[ \cos 247.5{}^\circ +i\sin 247.5{}^\circ \right]\text{ and} \\ & 2\left[ \cos 337.5{}^\circ +i\sin 337.5{}^\circ \right] \\ \end{align} Solutions of the provided equation in rectangular form are: $0.7654+1.8478i,-1.8478+0.7654i,-0.7654-1.8478i\text{ and }1.8478-0.7654i$
We apply Demoivre’s Theorem. Demoivre’s Theorem for finding complex roots states that every complex number has two distinct complex square roots, three distinct complex cube roots, four distinct complex fourth roots, and so on. Each root has the same modulus \begin{align} & r=z_{k}^{n} \\ & {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +360{}^\circ k}{n} \right)+i\sin \left( \frac{\theta +360{}^\circ k}{n} \right) \right] \\ \end{align} Here, $k$ is the number of distinct ${{n}^{th}}$ solutions and $\theta$ is in degrees. There are exactly four solutions of the provided equation as the degree of the equation is $4$ . \begin{align} & {{x}^{4}}=-16i \\ & \sqrt[4]{{{x}^{4}}}=\sqrt[4]{-16i} \\ & x=\sqrt[4]{-16i} \\ & =\sqrt[4]{16\left( \cos 270{}^\circ +i\sin 270{}^\circ \right)} \end{align} Apply Demoivre’s Theorem to find all solutions of the provided equation. Here, $n=4$ as the number of solutions is four. ${{z}_{k}}=\sqrt[4]{16}\left[ \cos \left( \frac{270{}^\circ +360{}^\circ k}{4} \right)+i\sin \left( \frac{270{}^\circ +360{}^\circ k}{4} \right) \right]$ Here, $k=0,1,2,3$. Insert above values of $k$ to find four distinct solution of the provided equation. For $k=0$ \begin{align} & {{z}_{0}}=\sqrt[4]{16}\left[ \cos \left( \frac{270{}^\circ +360{}^\circ \cdot \left( 0 \right)}{4} \right)+i\sin \left( \frac{270{}^\circ +360{}^\circ \cdot \left( 0 \right)}{4} \right) \right] \\ & =2\left[ \cos 67.5{}^\circ +i\sin 67.5{}^\circ \right] \end{align} Above solution is in the polar form. Substitute the value of $\cos 67.5{}^\circ$ and $\sin 67.5{}^\circ$ to find solution in rectangular form \begin{align} & {{z}_{0}}=\sqrt[4]{16}\left[ 0.3827+0.9239i \right] \\ & \approx 0.7654+1.8478i \end{align} For $k=1$ \begin{align} & {{z}_{1}}=\sqrt[4]{16}\left[ \cos \left( \frac{270{}^\circ +360{}^\circ \cdot \left( 1 \right)}{4} \right)+i\sin \left( \frac{270{}^\circ +360{}^\circ \cdot \left( 1 \right)}{4} \right) \right] \\ & =2\left[ \cos 157.5{}^\circ +i\sin 157.5{}^\circ \right] \end{align} Above solution is in the polar form. Substitute the value of $\cos 157.5{}^\circ$ and $\sin 157.5{}^\circ$ to find solution in rectangular form. \begin{align} & {{z}_{1}}=\sqrt[4]{16}\left[ \left( -0.9239 \right)+0.3827i \right] \\ & \approx -1.8478+0.7654i \end{align} For $k=2$ \begin{align} & {{z}_{2}}=\sqrt[4]{16}\left[ \cos \left( \frac{270{}^\circ +360{}^\circ \cdot \left( 2 \right)}{4} \right)+i\sin \left( \frac{270{}^\circ +360{}^\circ \cdot \left( 2 \right)}{4} \right) \right] \\ & =2\left[ \cos 247.5{}^\circ +i\sin 247.5{}^\circ \right] \end{align} Above solution is in the polar form. Substitute the value of $\cos 247.5{}^\circ$ and $\sin 247.5{}^\circ$ to find solution in rectangular form. \begin{align} & {{z}_{2}}=\sqrt[4]{16}\left[ -0.3827+\left( -0.9239 \right)i \right] \\ & \approx -0.7654-1.8478i \end{align} For $k=3$ \begin{align} & {{z}_{3}}=\sqrt[4]{16}\left[ \cos \left( \frac{270{}^\circ +360{}^\circ \cdot \left( 3 \right)}{4} \right)+i\sin \left( \frac{270{}^\circ +360{}^\circ \cdot \left( 3 \right)}{4} \right) \right] \\ & =2\left[ \cos 337.5{}^\circ +i\sin 337.5{}^\circ \right] \end{align} Above solution is in the polar form. Substitute the value of $\cos 337.5{}^\circ$ and $\sin 337.5{}^\circ$ to find solution in rectangular form. \begin{align} & {{z}_{3}}=\sqrt[4]{16}\left[ 0.9239+\left( -0.3827 \right)i \right] \\ & \approx 1.8478-0.7654i \end{align}