Cho số phức z =1+2i +3i^2 +...+2019i^2018 tìm z=?
$ z = (1 + 5i^4 + ... + 2017i^{2016}) + (3i^2 + 7i^6 + ... + 2019i^{2018}) + (2i + 6i^5 + ... + 2018i^{2017}) + (4i^3 + 8i^4 + ... + 2016i^{2015}) \\ = (1 + 5 . 1 + ... + 2017 . 1) + [3(-1) + 7(-1) + ... + 2019(-1)] + (2i + 6i + ... + 2018i) + [4(-i) + 8(-i) + ... + 2016(-i)] \\ = (1 + 5 + ... + 2017) - (3 + 7 + ... + 2019) + i(2 + 6 + ... + 2018) - i(4 + 8 + ... + 2016) \\ = \left (\frac{2017 - 1}{5 - 1} + 1 \right )\frac{1 + 2017}{2} - \left (\frac{2019 - 3}{7 - 3} + 1 \right )\frac{3 + 2019}{2} + i \left (\frac{2018 - 2}{6 - 2} + 1 \right )\frac{2 + 2018}{2} - i \left (\frac{2016 - 4}{8 - 4} + 1 \right )\frac{4 + 2016}{2} \\ = 505 . 1009 - 505 . 1011 + 505 . 1010i - 504 . 1010i \\ = 505 (1009 - 1011) + 1010i(505 - 504) \\ = 505 . (-2) + 1010i \\ = -1010 + 1010i $
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