+26393 find all the fourth roots of -8 in Cartesian form.
simplify as much as possilbe
\(\begin{array}{|rcll|} \hline \mathbf{z^4} & \mathbf{=} & \mathbf{-8 \qquad z_1,\ z_2,\ z_3,\ z_4 =\ ?}\\\\ z^4 &=& (-2)\cdot 4 \quad &| \quad \pm\sqrt{} \\ z^2 &=& \pm 2\cdot \sqrt{ -2 } \\ z^2 &=& \pm 2\cdot \sqrt{ (-1) \cdot 2 } \\ z^2 &=& \pm 2\cdot \sqrt{2} \cdot \sqrt{ -1 } \quad &| \quad \sqrt{ -1 } = i\\\\ \mathbf{z^2} & \mathbf{=} & \mathbf{\pm 2\cdot \sqrt{2} \cdot i }\\ \hline \end{array}\)
Solutions \(z_1,\ z_2,\ z_3,\ z_4 \):
\(\begin{array}{|lrcll|} \hline (1) & z_1^2 &=& +2\cdot \sqrt{2} \cdot i \\ (2) & z_2^2 = (-z_1)^2 &=& +2\cdot \sqrt{2} \cdot i \qquad \text{or} \qquad z_2 = -z_1\\ (3) & z_3^2 &=& -2\cdot \sqrt{2} \cdot i \\ (4) & z_4^2 = (-z_3)^2 &=& -2\cdot \sqrt{2} \cdot i \qquad \text{or} \qquad z_4 = -z_3\\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline \mathbf{z_1} \\\\ (1) & z_1^2 &=& +2\cdot \sqrt{2} \cdot i \quad &| \quad z_1 =a+b\cdot i \\ & (a+b\cdot i)^2 &=& +2\cdot \sqrt{2} \cdot i \\ & a^2 + 2abi + b^2i^2 &=& +2\cdot \sqrt{2} \cdot i \quad &| \quad i^2 = -1\\ & a^2 + 2abi - b^2 &=& +2\cdot \sqrt{2} \cdot i \\ & \underbrace{a^2 - b^2}_{Re=0} + \underbrace{2abi}_{Im=+2\cdot \sqrt{2} \cdot i} &=& +2\cdot \sqrt{2} \cdot i \\\\ & a^2-b^2 &=& 0 \\ & \mathbf{a^2} & \mathbf{=} & \mathbf{b^2}\\\\ & 2abi &=& +2\sqrt{2}\cdot i \\ & ab &=& \sqrt{2} \\ & \mathbf{b} &\mathbf{=}& \mathbf{ \frac{ \sqrt{2} } {a} } \\\\ & a^2 &=& b^2 \\ & a^2 &=& (\frac{ \sqrt{2} } {a} )^2 \\ & a^2 &=& \frac{ 2 } {a^2} \\ & a^4 &=& 2 \\ & \mathbf{a} &\mathbf{=}& \mathbf{\sqrt[4]{2}} \\\\ & b &=& \frac{ \sqrt{2} } {a} \\ & b &=& \frac{ \sqrt{2} } {\sqrt[4]{2}} \\ & b &=& \frac{ 2^{\frac12} } { 2^{\frac14} } \\ & b &=& 2^{\frac12-\frac14} \\ & b &=& 2^{\frac14} \\ & \mathbf{b} &\mathbf{=}& \mathbf{\sqrt[4]{2}} \\\\ & z_1 &=& a+bi \\ & z_1 &=& \sqrt[4]{2} + \sqrt[4]{2} \cdot i \\ & \mathbf{z_1} &\mathbf{=}& \mathbf{\sqrt[4]{2}\cdot (1 + i)} \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline \mathbf{z_2} \\\\ (2) & z_2 &=& -z_1 \\ & \mathbf{z_2} &\mathbf{=}& \mathbf{-\sqrt[4]{2}\cdot (1 + i)} \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline \mathbf{z_3} \\\\ (3) & z_3^2 &=& -2\cdot \sqrt{2} \cdot i \quad &| \quad z_3 =a+b\cdot i \\ & (a+b\cdot i)^2 &=& -2\cdot \sqrt{2} \cdot i \\ & a^2 + 2abi + b^2i^2 &=& -2\cdot \sqrt{2} \cdot i \quad &| \quad i^2 = -1\\ & a^2 + 2abi - b^2 &=& -2\cdot \sqrt{2} \cdot i \\ & \underbrace{a^2 - b^2}_{Re=0} + \underbrace{2abi}_{Im=-2\cdot \sqrt{2} \cdot i} &=& -2\cdot \sqrt{2} \cdot i \\\\ & a^2-b^2 &=& 0 \\ & \mathbf{a^2} & \mathbf{=} & \mathbf{b^2}\\\\ & 2abi &=& -2\sqrt{2}\cdot i \\ & ab &=& -\sqrt{2} \\ & \mathbf{b} &\mathbf{=}& \mathbf{ \frac{ -\sqrt{2} } {a} } \\\\ & a^2 &=& b^2 \\ & a^2 &=& (\frac{ -\sqrt{2} } {a} )^2 \\ & a^2 &=& \frac{ 2 } {a^2} \\ & a^4 &=& 2 \\ & \mathbf{a} &\mathbf{=}& \mathbf{\sqrt[4]{2}} \\\\ & b &=& \frac{ -\sqrt{2} } {a} \\ & b &=& \frac{ -\sqrt{2} } {\sqrt[4]{2}} \\ & b &=& -\frac{ 2^{\frac12} } { 2^{\frac14} } \\ & b &=& -2^{\frac12-\frac14} \\ & b &=& -2^{\frac14} \\ & \mathbf{b} &\mathbf{=}& \mathbf{-\sqrt[4]{2}} \\\\ & z_3 &=& a+bi \\ & z_3 &=& \sqrt[4]{2} - \sqrt[4]{2} \cdot i \\ & \mathbf{z_3} &\mathbf{=}& \mathbf{\sqrt[4]{2}\cdot (1 - i)} \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline \mathbf{z_4} \\\\ (4) & z_4 &=& -z_3 \\ & \mathbf{z_4} &\mathbf{=}& \mathbf{-\sqrt[4]{2}\cdot (1 - i)} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{z_1} &\mathbf{=}& \mathbf{\sqrt[4]{2}\cdot (1 + i)} \\ \mathbf{z_2} &\mathbf{=}& \mathbf{-\sqrt[4]{2}\cdot (1 + i)} \\ \mathbf{z_3} &\mathbf{=}& \mathbf{\sqrt[4]{2}\cdot (1 - i)} \\ \mathbf{z_4} &\mathbf{=}& \mathbf{-\sqrt[4]{2}\cdot (1 - i)} \\ \hline \end{array}\)
