# My work with complex numbers verified that the only possible

My work with complex numbers verified that the only possible cube root of 8 is 2. Determine whether the statement makes sense or does not make sense, and explain your reasoning.
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Step 1
Concept:
The calculus helps in understanding the changes between values that are related by a function. It is used in different areas such as physics, biology and sociology and so on. The calculus is the language of engineers, scientists and economics.
Step 2
Given:
It is given that,''My work with complex numbers verified that the only possible cube root of 8 is 2''
The objective is to check that the sentence make sense or not
Step 3
This given statement does not makes sense
Here DeMoivre's theorem for roots applies to complex numbers in polar form. Thus writing the given term in terms of polar form that is 8 or $8+0i$ in the polar form Expressing $\theta$ in radians although degrees could also be used
$8=r\left(\mathrm{cos}\theta +i\mathrm{sin}\theta \right)=8\left(\mathrm{cos}0+i\mathrm{sin}0\right)$
There is three cube root exactly of 8. From DeMoivre's theorem for finding complex roots, the cube roots of 8 are
${Z}_{k}=\sqrt[3]{8}\left(\mathrm{cos}\left(\frac{0+2\pi k}{3}\right)+i\mathrm{sin}\left(\frac{0+2\pi k}{3}\right)\right)$

Step 4
The cube roots of 8 are found by substituting 0, 1, 2 for k in the above expression
${Z}_{0}=\sqrt[3]{8}\left(\mathrm{cos}\left(\frac{2\pi 0}{3}\right)+i\mathrm{sin}\left(\frac{2\pi 0}{3}\right)\right)$
${Z}_{0}=2\left({\mathrm{cos}0}^{0}+i{\mathrm{sin}0}^{0}\right)⇒$ Polar form
${Z}_{0}=2⇒$ rectangular form
${Z}_{1}=2\left(\mathrm{cos}\left(\frac{2\pi }{3}\right)+i\mathrm{sin}\left(\frac{2\pi }{3}\right)\right)$
${Z}_{1}=2\left({\mathrm{cos}120}^{\circ }+i{\mathrm{sin}120}^{\circ }\right)$
Thus polar form,
${Z}_{1}=2\left({\mathrm{cos}120}^{\circ }+i{\mathrm{sin}120}^{\circ }\right)$
The rectangular form
${Z}_{1}=2\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)$
Step 5
${Z}_{2}=\sqrt[3]{8}\left(\mathrm{cos}\left(\frac{4\pi }{3}{\right)}_{i}\mathrm{sin}\left(\frac{4\pi }{3}\right)\right)$
${Z}_{2}=2\left({\mathrm{cos}240}^{\circ }+i{\mathrm{sin}240}^{\circ }\right)$
polar form
${Z}_{2}=2\left(-{\mathrm{cos}60}^{\circ }-i{\mathrm{sin}60}^{\circ }\right)$
Rectangular form
${Z}_{2}=-2\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)$