[amp_mcq option1=”j” option2=”$$\\frac{{\\sqrt 3 }}{2} + {\\text{j}}\\frac{1}{2}$$” option3=”$$\\frac{{\\sqrt 3 }}{2} – {\\text{j}}\\frac{1}{2}$$” option4=”$$ – \\frac{{\\sqrt 3 }}{2} – {\\text{j}}\\frac{1}{2}$$” correct=”option1″]<\/p>\n
The correct answer is $\\boxed{\\frac{{\\sqrt 3 }}{2} + {\\text{j}}\\frac{1}{2}}$.<\/p>\n
To solve for the roots of the equation $x^3 = j$, we can use the cubic formula:<\/p>\n
$$x = \\dfrac{-b \\pm \\sqrt{b^2 – 4ac}}{2a}$$<\/p>\n
where $a = 1$, $b = 0$, and $c = -j$.<\/p>\n
Substituting these values into the formula, we get:<\/p>\n
$$x = \\dfrac{0 \\pm \\sqrt{0^2 – 4 \\cdot 1 \\cdot -j}}{2 \\cdot 1}$$<\/p>\n
$$x = \\dfrac{0 \\pm \\sqrt{4j}}{2}$$<\/p>\n
$$x = \\dfrac{0 \\pm 2\\sqrt{j}}{2}$$<\/p>\n
$$x = \\pm \\sqrt{j}$$<\/p>\n
Since $j$ is the positive square root of $-1$, then $\\sqrt{j} = j$. Therefore, the roots of the equation $x^3 = j$ are $x = \\pm j$.<\/p>\n
We can also solve for the roots by inspection. Note that $j$ is a complex number with magnitude $1$ and angle $\\frac{\\pi}{3}$. Therefore, the roots of the equation $x^3 = j$ must be complex numbers with magnitude $1$ and angles $\\frac{\\pi}{3}$, $\\frac{2\\pi}{3}$, and $\\frac{4\\pi}{3}$. The only complex number with magnitude $1$ and angle $\\frac{\\pi}{3}$ is $\\frac{{\\sqrt 3 }}{2} + {\\text{j}}\\frac{1}{2}$. Therefore, $\\boxed{\\frac{{\\sqrt 3 }}{2} + {\\text{j}}\\frac{1}{2}}$ is one of the roots of the equation $x^3 = j$.<\/p>\n
The other two roots of the equation $x^3 = j$ are $-\\frac{{\\sqrt 3 }}{2} – {\\text{j}}\\frac{1}{2}$ and $\\frac{{\\sqrt 3 }}{2} – {\\text{j}}\\frac{1}{2}$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”j” option2=”$$\\frac{{\\sqrt 3 }}{2} + {\\text{j}}\\frac{1}{2}$$” option3=”$$\\frac{{\\sqrt 3 }}{2} – {\\text{j}}\\frac{1}{2}$$” option4=”$$ – \\frac{{\\sqrt 3 }}{2} – {\\text{j}}\\frac{1}{2}$$” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20250","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n