91Ó°ÊÓ

For any complex numbers, let say,$a\text{and}b$ the definition of the complex power induces a formula as: $a^b=e^{blna}$, where.$a≠e$

The given polynomial is$x^3−3x−1=0$, with roots .$2\cos \left(\frac{\mathrm{Ï€}}{9}\right),\text{ }−2\cos \left(\frac{2\mathrm{Ï€}}{9}\right),\text{²¹²Ô»å }−2\cos \left(\frac{4\mathrm{Ï€}}{9}\right)$

Using computer, the three roots obtained are:

$\begin{array}{c}{x}_1=1.88\\ x_2=−0.347\\ x_3=−1.53\end{array}$

Let these roots are real part of the complex roots$z_1,z_2\text{and}{z}_3$given by:

$\begin{array}{l}{z}_1=x_1+i{y}_1=2\{\cos {\mathrm{θ}}_1+i\sin {\mathrm{θ}}_1\}\\ z_2=x_2+i{y}_2=2\{\cos {\mathrm{θ}}_2+i\sin {\mathrm{θ}}_2\}\\ z_3=x_3+i{y}_3=2\{\cos {\mathrm{θ}}_3+i\sin {\mathrm{θ}}_3\}\end{array}$

From the equation for $z_1$solve for the roots as:

$\begin{array}{c}{x}_1=2\cos {\mathrm{θ}}_1=1.88\\ {\mathrm{θ}}_1={\cos }^{-1}\left[\frac{1.88}{2}\right]=Â\pm \frac{\mathrm{Ï€}}{9}\\ \mathrm{Þ}{x}_1=2\cos \left(\frac{\mathrm{Ï€}}{9}\right)\end{array}$

From the equation for$z_2$solve for the roots as:

$\begin{array}{c}{x}_2=2\cos {\mathrm{θ}}_2=−0.347\\ {\mathrm{θ}}_2={\cos }^{−1}\left[\frac{0.347}{2}\right]=Â\pm \frac{4\mathrm{Ï€}}{9}\\ ⇒x_2=−2\cos \left(\frac{4\mathrm{Ï€}}{9}\right)\end{array}$

From the equation for$z_3$solve for the roots as:

$\begin{array}{c}{x}_3=2\cos {\mathrm{θ}}_3=−1.53\\ {\mathrm{θ}}_3={\cos }^{−1}\left[\frac{1.53}{2}\right]=Â\pm \frac{2\mathrm{Ï€}}{9}\\ ⇒x_3=−2\cos \left(\frac{2\mathrm{Ï€}}{9}\right)\end{array}$

Hence, the solutions can be written as:

$\begin{array}{c}{x}_1=1.88=2\cos \left(\frac{\mathrm{π}}{9}\right)\\ x_2=−0.347=−2\cos \left(\frac{4\mathrm{π}}{9}\right)\\ x_3=−1.53=−2\cos \left(\frac{2\mathrm{π}}{9}\right)\end{array}$