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Sir pls solve the following.
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Asked by rsudipto | 28 Dec, 2018, 07:54: PM
answered-by-expert Expert Answer
begin mathsize 16px style straight z equals open parentheses fraction numerator square root of 3 over denominator 2 end fraction plus straight i over 2 close parentheses to the power of 5 plus open parentheses fraction numerator square root of 3 over denominator 2 end fraction minus straight i over 2 close parentheses to the power of 5 straight z equals open parentheses cos straight pi over 6 plus isin straight pi over 6 close parentheses to the power of 5 plus open parentheses cos straight pi over 6 minus isin straight pi over 6 close parentheses to the power of 5 straight z equals cos fraction numerator 5 straight pi over denominator 6 end fraction plus straight i sin fraction numerator 5 straight pi over denominator 6 end fraction plus cos fraction numerator 5 straight pi over denominator 6 end fraction minus isin fraction numerator 5 straight pi over denominator 6 end fraction straight z equals 2 cos fraction numerator 5 straight pi over denominator 6 end fraction straight z equals cos open parentheses straight pi minus straight pi over 6 close parentheses straight z equals 2 cross times fraction numerator negative square root of 3 over denominator 2 end fraction straight z equals negative square root of 3 Re less than 0 Im space equals space 0 end style
Answered by Sneha shidid | 31 Dec, 2018, 11:48: AM

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JEE main - Maths
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