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Question Number 70704 by 20190927 last updated on 07/Oct/19

$${\mathrm{z}}^4-{2\mathrm{z}}^2+2=0$$

Commented by mathmax by abdo last updated on 08/Oct/19

$$z^4-2z^2+2=0\Rightarrow t^2-2t+2=0witht=z^2$$$${\mathrm{\Delta }}^{{}^{\prime }}=1-2=-1\Rightarrow t_1=1+iand{t}_2=1-i$$$$t_1=\sqrt{2}{e}^{\frac{i\pi }{4}}and{t}_2=\sqrt{2}{e}^{-\frac{i\pi }{4}}$$$$z^2=t_1\Rightarrow z=\stackrel{-}{+}\sqrt{t_1}=\stackrel{-}{+}\left({}^4\sqrt{2}\right)e^{\frac{i\pi }{8}}$$$$z^2=t_2\Rightarrow z=\stackrel{-}{+}\sqrt{t_2}=\stackrel{-}{+}\left({}^4\sqrt{2}\right)e^{-\frac{i\pi }{8}}sotherootsare\stackrel{-}{+}\left({}^4\sqrt{2}\right)e^{\frac{i\pi }{8}}$$$$and\stackrel{-}{+}\left({}^4\sqrt{2}\right)e^{-\frac{i\pi }{8}}alsothefactorizationis$$$$z^4-2z^2+2=(z-\left({}^4\sqrt{2}\right)e^{\frac{i\pi }{8}})\left(z{+}^4\sqrt{2}{e}^{\frac{i\pi }{8}}\right)\left(z{-}^4\sqrt{2}{e}^{-\frac{i\pi }{8}}\right)\left(z{+}^4\sqrt{2}{e}^{-\frac{i\pi }{8}}\right)$$

Commented by 20190927 last updated on 09/Oct/19

$$\mathrm{thank}\mathrm{you}$$

Answered by Rasheed.Sindhi last updated on 07/Oct/19

$${\mathrm{z}}^4-{2\mathrm{z}}^2+2=0$$$${\left(z^2\right)}^2-2z^2+2=0$$$$z^2=y\Rightarrow y^2-2y+2=0$$$$y=\frac{-(-2)\pm \sqrt{{(-2)}^2-4\left(1\right)\left(2\right)}}{2\left(1\right)}$$$$=\frac{2\pm \sqrt{(4-8}}{2}=\frac{2\pm \sqrt{-4}}{2}=\frac{2\pm 2i}{2}$$$$y=1\pm i$$$$z^2=1\pm i$$$$z=\pm \sqrt{1\pm i}$$

Commented by 20190927 last updated on 09/Oct/19

$$\mathrm{thank}\mathrm{you}$$

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