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If to ω ≠ 1 is a cube root of unity, then show that abcbcaabccaaa+bω+cω2b+cω+aω2+a+bω+cω2c+aω+aω2 = – 1 - Mathematics

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प्रश्न

If to ω ≠ 1 is a cube root of unity, then show that `("a" + "b"omega + "c"omega^2)/("b" + "c"omega + "a"omega^2) + ("a" + "b"omega + "c"omega^2)/("c" + "a"omega + "a"omega^2)` = – 1

योग

उत्तर

L.H.S

`("a" + "b"omega + "c"omega^2)/("b" + "c"omega + "a"omega^2) + ("a" + "b"omega + "c"omega^2)/("c" + "a"omega + "a"omega^2)`

= `(("a" + "b"omega + "c"omega^2)/("b" + "c"omega + "a"omega^2)) omega/omega + (("a" + "b"omega + "c"omega^2)/("c" + "a"omega + "b"omega^2)) omega^2/omega^2`

= `(("a" + "b"omega + "c"omega^2)omega)/("b"omega + "c"omega^2 + "a"omega^3) + (("a" + "b"omega + "c"omega^2)omega^2)/("c"omega^2 + "a"omega^3 + "b"omega^4)`

= `(("a" + "b"omega + "c"omega^2)omega)/(("b"omega + "c"omega^2 + "a")) + (("a" + "b"omega + "c"omega^2)omega^2)/(("c"omega^2 + "a" + "b"omega))`  ......`[because omega^3 = 1, omega^4 = omega^3 * omega = omega]`

= `omega + omega^2`  ......`[because 1 + omega + omega^2 = 0]`

= – 1

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de Moivre’s Theorem and Its Applications
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Complex Numbers - Exercise 2.8 [पृष्ठ ९२]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 2 Complex Numbers
Exercise 2.8 | Q 1 | पृष्ठ ९२

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