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Question Number 158274 by HongKing last updated on 01/Nov/21
Solve for complex numbers:  x^4  + (1 + i)x^3  + 2ix^2  + (i - 1)x - 1 = 0
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{complex}\:\mathrm{numbers}: \\ $$$$\mathrm{x}^{\mathrm{4}} \:+\:\left(\mathrm{1}\:+\:\boldsymbol{\mathrm{i}}\right)\boldsymbol{\mathrm{x}}^{\mathrm{3}} \:+\:\mathrm{2}\boldsymbol{\mathrm{ix}}^{\mathrm{2}} \:+\:\left(\boldsymbol{\mathrm{i}}\:-\:\mathrm{1}\right)\boldsymbol{\mathrm{x}}\:-\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$ \\ $$
Answered by mindispower last updated on 01/Nov/21
x=−1, solution   (x+1)(x^3 +ix^2 +ix−1)  (x+1)(x+i)(x^2 +i)  (x+1)(x+i)(x−e^(−((iπ)/4)) )(x−e^((3iπ)/4) )
$${x}=−\mathrm{1},\:{solution}\: \\ $$$$\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{3}} +{ix}^{\mathrm{2}} +{ix}−\mathrm{1}\right) \\ $$$$\left({x}+\mathrm{1}\right)\left({x}+{i}\right)\left({x}^{\mathrm{2}} +{i}\right) \\ $$$$\left({x}+\mathrm{1}\right)\left({x}+{i}\right)\left({x}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({x}−{e}^{\frac{\mathrm{3}{i}\pi}{\mathrm{4}}} \right) \\ $$

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