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Many of you might have the opinion that only \u20181\u2019 is the cube root of \u20181\u2019 but no, that\u2019s not the case. There are other cube roots of \u20181\u2019 also. In this article, we are going to learn about the meaning, properties and some illustrations of the cube roots of unity.\u00a0<\/span><\/p>\n <\/strong><\/p>\n When a number is raised to the power of 3 and it results in getting the number \u20181\u2019, then it is said to be a cube of unity and the inverse of it is the cube root of unity. In simple words, we need to determine those numbers whose cube will give the result \u20181\u2019. The cube roots of unity are 1, \\frac{-1+\\sqrt{3} i}{2}, \\frac{-1-\\sqrt{3} i}{2}<\/span>. Here \u20181\u2019 is the real root and <\/span><\/p>\n \\frac{-1+\\sqrt{3 i}}{2} \\& \\frac{-1-\\sqrt{3 i}}{2}<\/span> are imaginary roots.<\/span><\/p>\n <\/p>\n Suppose \u2018t\u2019 is the number whose cube is 1.<\/span><\/p>\n \\text { So, } t^{3}=1, \\quad t^{3}-1^{3}=0<\/span><\/span><\/p>\n (t-1)\\left(t^{2}+t+1\\right)=0 \\quad\\left[a^{3}-b^{3}=(a-b)\\left(a^{2}+a b+b^{2}\\right)\\right]<\/span><\/span><\/p>\n t-1=0, \\quad t^{2}+t+1=0<\/span><\/span><\/p>\n \u00a0So, t = 1<\/span><\/p>\n For the equation t^{2}+t+1=0<\/span><\/span>, we will use the quadratic roots formula.<\/span><\/p>\n We know that the formula to determine the root is\u00a0 \u00a0 x=\\frac{-b \\pm \\sqrt{d}}{2 a}<\/span><\/span>, \u00a0 <\/span>where d=b^{2}-4 a c<\/span>.<\/span><\/p>\n The values of a, b and c here are 1, 1 and 1 respectively. So, d=1^{2}-4=-3<\/span>.<\/span><\/p>\n Therefore, t=\\frac{-1 \\pm \\sqrt{-3}}{2}<\/span><\/span><\/p>\n We know that the value \u221a-1 = iota(i<\/span>)<\/span><\/p>\n Hence, the two imaginary roots of unity are \\frac{-1+\\sqrt{3} i}{2} \\& \\frac{-1-\\sqrt{3} i}{2} \\text { where } \\omega=\\frac{-1+\\sqrt{3} i}{2} \\text { and } \\omega^{2}=\\frac{-1-\\sqrt{3} i}{2} .<\/span><\/span><\/p>\n <\/p>\n The three cube roots are 1 which is a real number and \\frac{-1+\\sqrt{3} i}{2} \\& \\frac{-1-\\sqrt{3} i}{2}<\/span><\/span>are conjugate complex or imaginary numbers.<\/span><\/p>\n <\/span><\/p>\n \u00a0It can be proved with the help of cube roots. We know that the cube roots of unity are 1,<\/span><\/p>\n\\frac{-1+\\sqrt{3 i}}{2}, \\frac{-1-\\sqrt{3 i}}{2}<\/span>\n \\left(\\frac{-1+\\sqrt{3} i}{2}\\right)^{2}=\\frac{(-1)^{2}+(\\sqrt{3} i)^{2}-2 \\sqrt{3} i}{2^{2}}<\/span> It means that <\/span>\u03c9 =\\frac{-1+\\sqrt{3 i}}{2}<\/span><\/span>and<\/span> {\\omega}^2=\\frac{-1-\\sqrt{3 i}}{2}<\/span><\/span>, 1, \u03c9 and {\\omega}^2<\/span><\/span>\u00a0are the cube roots of unity.<\/span><\/p>\n Hence, it is proved that<\/span> the square of one complex cube root of unity is equal to the other complex cube root of unity.\u00a0<\/span><\/p>\n <\/p>\n We can verify it, the three cube roots of unity are<\/span> 1, \\frac{-1+\\sqrt{3 i}}{2}, \\frac{-1-\\sqrt{3 i}}{2} .<\/span><\/span><\/p>\n So, 1 \\times \\frac{-1+\\sqrt{3 i}}{2} \\times \\frac{-(1+\\sqrt{3 i})}{2}<\/span><\/span><\/p>\n=\\frac{\\left.-(\\sqrt{3} i)^{2}-(1)^{2}\\right)}{2^{2}} \\quad\\left[(a+b)(a-b)=a^{2}-b^{2}\\right]<\/span>\n <\/p>\n=\\frac{-(-3-1)}{4}<\/span>\n <\/p>\n = \\frac{4}{4}= 1 <\/span><\/span><\/p>\n We know that the product of cube roots is equal to 1.\u00a0<\/span><\/p>\n Therefore, \\omega^{3}=1<\/span> <\/span>&<\/span> the value of <\/span>\u2375 <\/span>is 1.<\/span><\/p>\n <\/span><\/p>\n We know that, the sum of the three cube roots of unity =1+\\omega+\\omega^{2}<\/span><\/span>\u00a0<\/span><\/p>\n So, 1+\\frac{-1+\\sqrt{3 i}}{2}+\\frac{-1-\\sqrt{3 i}}{2}<\/span><\/span><\/p>\n Taking LCM,<\/p>\n We will get \\frac{2-1+\\sqrt{3 i}-1-\\sqrt{3 i}}{2}<\/span><\/span>, which is equal to 0.<\/span><\/p>\n <\/span><\/p>\n <\/p>\n Q1: Find the value of the following :\\left(1+\\omega-\\omega^{2}\\right)\\left(1-\\omega+\\omega^{2}\\right)<\/span><\/span>\u00a0\u00a0<\/span><\/p>\n Sol. We need to do the multiplication to find its value,<\/span><\/p>\n =\\left(1+\\omega-\\omega^{2}\\right)\\left(1-\\omega+\\omega^{2}\\right) <\/span> <\/span><\/p>\n Q2. If (1+\\omega)^{7}=C+D \\omega <\/span>, then find the value of C and D using the cube root of unity. (\u2375 = 1)<\/span><\/p>\n Sol. <\/strong><\/p>\n Given: (1+\\omega)^{7}=C+D \\omega<\/span><\/span><\/p>\n (1+\\omega)^{6}(1+\\omega)=C+D \\omega <\/span><\/span><\/p>\n {\\left[(1+\\omega)^{(2)^{3}}\\right](1+\\omega)=C+D \\omega}<\/span><\/span><\/p>\n (1+\\omega^2 +2\\omega)^{3}(1+\\omega )=C+D\\omega<\/span><\/span><\/span><\/p>\n (-\\omega+2 \\omega)^{3}(1+\\omega)=C+D \\omega <\/span><\/span><\/p>\n \\omega^{3}(1+\\omega)=C+D \\omega\u00a0 \u00a0 [\\text{Since }\\omega^{3}=1]<\/span><\/span><\/p>\n <\/p>\n By comparing LHS and RHS,<\/span><\/p>\n The value of C and D is 1 each.<\/span><\/p>\n [\/et_pb_text][\/et_pb_column][et_pb_column type=”2_5″ module_id=”stickysideR” admin_label=”Column R” _builder_version=”4.10.4″ _module_preset=”default” background_color=”#fdefe0″ custom_padding=”25px|25px|25px|25px|true|true” sticky_position=”top” sticky_offset_top=”-280px” sticky_limit_top=”row” sticky_limit_bottom=”row” sticky_position_tablet=”none” sticky_position_phone=”none” sticky_position_last_edited=”on|desktop” sticky_limit_bottom_tablet=”” sticky_limit_bottom_phone=”” sticky_limit_bottom_last_edited=”on|phone” border_radii=”on|15px|15px|15px|15px” box_shadow_style=”preset3″ global_colors_info=”{}”][et_pb_image src=”https:\/\/eistudymaterial.s3.amazonaws.com\/1080×1080.png” alt=”Free Trial banner” title_text=”Mindspark Free Trial Banner” url=”https:\/\/mindspark.in\/free-trial” align=”center” module_class=”adsimg” _builder_version=”4.11.1″ _module_preset=”default” custom_padding=”||||false|false” border_radii=”on|10px|10px|10px|10px” global_colors_info=”{}” transform_styles__hover_enabled=”on|hover” transform_scale__hover_enabled=”on|hover” transform_translate__hover_enabled=”on|desktop” transform_rotate__hover_enabled=”on|desktop” transform_skew__hover_enabled=”on|desktop” transform_origin__hover_enabled=”on|desktop” transform_scale__hover=”102%|102%”][\/et_pb_image][et_pb_text admin_label=”Explore Other Topics [\/et_pb_text][et_pb_text _builder_version=”4.10.7″ _module_preset=”default” text_line_height=”2.2em” link_font_size=”16px” custom_margin=”||0px||false|false” custom_padding=”10px|15px|10px|28px|true|false” locked=”off” global_colors_info=”{}”]<\/p>\nCUBE ROOTS OF UNITY<\/strong><\/span><\/h2>\n
MEANING\u00a0<\/span><\/h3>\n
HOW DO YOU FIND THE ROOTS OF UNITY?<\/strong><\/h2>\n
PROPERTIES OF THE CUBE ROOT OF UNITY<\/strong><\/h2>\n
<\/h3>\n
Property 1: <\/span>\u00a0<\/span>There are three cube roots of unity of which one is a real number and the other two are conjugate complex or imaginary numbers.<\/span><\/h3>\n
Property 2: Square of one complex number cube root of unity is equal to the other complex number cube root of unity.<\/span><\/h3>\n
=\\frac{1+3 i^{2}-2 \\sqrt{3} i}{4} <\/span>
=\\frac{1-3-2 \\sqrt{3} i}{4} <\/span>
=\\frac{2(-1-\\sqrt{3} i)}{4} <\/span>
=\\frac{-1-\\sqrt{3} i}{2}<\/span><\/p>\nProperty 3:The product of the two imaginary cube roots is 1 or, the product of three cube roots of unity is 1.<\/span><\/h3>\n
Property 4: The addition of all the cube roots of unity is zero i.e.,1+\\omega+\\omega^{2}=0<\/span><\/span><\/h3>\n
ILLUSTRATION<\/h2>\n
=1-\\omega+\\omega^{2}+\\omega-\\omega^{2}+\\omega^{3}-\\omega^{2}+\\omega^{3}-\\omega^{4} <\/span>
=1-\\omega^{2}-\\omega^{4}+2 \\omega^{3}<\/span>
=1+2-\\omega^{2}\\left(1+\\omega^{2}\\right) \\quad\\left[\\omega^{3}=1\\right]<\/span>
=3-\\omega^{2}(-\\omega) \\quad\\left[1+\\omega+\\omega^{2}=0,1+\\omega^{2}=-\\omega\\right]<\/span>
=3+\\omega^{3} <\/span>
=4<\/span><\/span><\/span><\/p>\n
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