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Question Number 52498 by Joel578 last updated on 09/Jan/19

$$\mathrm{Let}z\mathrm{is}\mathrm{complex}\mathrm{number}\mathrm{and}\mathrm{satisfy}{z}^{2017}=1,z\ne 1$$$$\mathrm{Find}(1+z)(1+z^2)(1+z^3)\dots (1+z^{2016})$$

Commented by mr W last updated on 09/Jan/19

$$P=(1+z)(1+z^2)(1+z^3)\dots (1+z^{2016})$$$$=1+z+z^2+...+z^{1+2+3+...+2016}$$$$=1+z+z^2+...+z^{\frac{2016\times 2017}{2}}$$$$=1+z+z^2+...+z^{1008\times 2017}$$$$=\frac{1-z^{1008\times 2017+1}}{1-z}$$$$=\frac{1-z{\left(z^{2017}\right)}^{1008}}{1-z}$$$$=\frac{1-z\times 1^{1008}}{1-z}$$$$=\frac{1-z}{1-z}$$$$=1$$

Answered by MJS last updated on 09/Jan/19

$$z^{2n+1}=1,z\ne 1:\underset{k=1}{\prod ^{2n}}(1+z^k)=1$$

Commented by malwaan last updated on 10/Jan/19

$$\mathrm{how}?$$

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