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Question Number 73525 by arkanmath7@gmail.com last updated on 13/Nov/19

Commented by mathmax by abdo last updated on 13/Nov/19

$$z^2+(1-i)z-3i=0$$$$\mathrm{\Delta }={(1-i)}^2-4(-3i)=1-2i-1+12i=10i$$$$z_1=\frac{-1+i+\sqrt{10}{e}^{\frac{i\pi }{4}}}{2}and{z}_2=\frac{-1+i+\sqrt{10}{e}^{\frac{i\pi }{4}}}{2}and$$$$z^2+(1-i)z-3i=(z-\frac{-1+i+\sqrt{10}{e}^{\frac{i\pi }{4}}}{2})(z-\frac{-1+i+\sqrt{10}{e}^{\frac{i\pi }{4}}}{2})$$

Commented by arkanmath7@gmail.com last updated on 15/Nov/19

$$thnx$$

Answered by MJS last updated on 13/Nov/19

$$z=-\frac{1-\mathrm{i}}{2}\pm \sqrt{\frac{{(1-\mathrm{i})}^2}{4}+3\mathrm{i}}=$$$$=-\frac{1}{2}+\frac{1}{2}\mathrm{i}\pm \sqrt{\frac{5\mathrm{i}}{2}}=$$$$=-\frac{1}{2}\pm \frac{\sqrt{5}}{2}+(\frac{1}{2}\pm \frac{\sqrt{5}}{2})\mathrm{i}$$$$z_1=-\frac{1+\sqrt{5}}{2}+\frac{1-\sqrt{5}}{2}\mathrm{i}$$$$z_2=-\frac{1-\sqrt{5}}{2}+\frac{1+\sqrt{5}}{2}\mathrm{i}$$$$(z+\frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}\mathrm{i})(z+\frac{1-\sqrt{5}}{2}-\frac{1+\sqrt{5}}{2}\mathrm{i})=0$$

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