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Asked by g_archanasharma | 18 Dec, 2018, 20:21: PM
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imaginary number begin mathsize 12px style i space equals space square root of negative 1 end root space equals space cos straight pi over 2 plus i space sin straight pi over 2 end style
begin mathsize 12px style cube root of negative 1 end root space equals space i to the power of bevelled 1 third end exponent space equals space open parentheses cos open parentheses straight pi over 2 plus 2 k straight pi close parentheses plus i space sin open parentheses straight pi over 2 plus 2 k straight pi close parentheses space close parentheses to the power of bevelled 1 third end exponent space equals space cos space fraction numerator open parentheses 4 k plus 1 close parentheses over denominator 6 end fraction straight pi space plus space i space sin fraction numerator open parentheses 4 k plus 1 close parentheses over denominator 6 end fraction straight pi space comma space w h e r e space k space equals space 0 comma 1 comma 2 w h e n space k equals space 0 space colon space space space i to the power of bevelled 1 third end exponent space equals space cos space straight pi over 6 space plus space i space sin straight pi over 6 space equals space fraction numerator square root of 3 over denominator 2 end fraction plus i 1 half space equals space fraction numerator square root of 3 space plus space i over denominator 2 end fraction w h e n space k space equals space 1 space colon space space space i to the power of bevelled 1 third end exponent space equals space cos space fraction numerator 5 straight pi over denominator 6 end fraction space plus space i space sin fraction numerator 5 straight pi over denominator 6 end fraction space equals space minus fraction numerator square root of 3 over denominator 2 end fraction plus i 1 half space equals space fraction numerator negative square root of 3 space plus space i over denominator 2 end fraction w h e n space k space equals space 2 space colon space space i to the power of bevelled 1 third end exponent space equals space cos space fraction numerator 3 straight pi over denominator 2 end fraction space plus space i space sin fraction numerator 3 straight pi over denominator 2 end fraction space equals space 0 minus i space equals space minus i end style
Hence (1) ans (4) are cube root of i which is √-1
Answered by Thiyagarajan K | 19 Dec, 2018, 13:04: PM

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