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Question-73525




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  Δ=(1−i)^2 −4(−3i) =1−2i−1+12i =10i  z_1 =((−1+i+(√(10))e^((iπ)/4) )/2)  and z_2  =((−1+i+(√(10))e^((iπ)/4) )/2)  and  z^2 +(1−i)z −3i =(z−((−1+i+(√(10))e^((iπ)/4) )/2))(z−((−1+i+(√(10))e^((iπ)/4) )/2))
$${z}^{\mathrm{2}} +\left(\mathrm{1}−{i}\right){z}−\mathrm{3}{i}\:=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{1}−{i}\right)^{\mathrm{2}} −\mathrm{4}\left(−\mathrm{3}{i}\right)\:=\mathrm{1}−\mathrm{2}{i}−\mathrm{1}+\mathrm{12}{i}\:=\mathrm{10}{i} \\ $$$${z}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}+\sqrt{\mathrm{10}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{2}}\:\:{and}\:{z}_{\mathrm{2}} \:=\frac{−\mathrm{1}+{i}+\sqrt{\mathrm{10}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{2}}\:\:{and} \\ $$$${z}^{\mathrm{2}} +\left(\mathrm{1}−{i}\right){z}\:−\mathrm{3}{i}\:=\left({z}−\frac{−\mathrm{1}+{i}+\sqrt{\mathrm{10}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{2}}\right)\left({z}−\frac{−\mathrm{1}+{i}+\sqrt{\mathrm{10}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{2}}\right) \\ $$
Commented by arkanmath7@gmail.com last updated on 15/Nov/19
thnx
$${thnx} \\ $$
Answered by MJS last updated on 13/Nov/19
z=−((1−i)/2)±(√((((1−i)^2 )/4)+3i))=  =−(1/2)+(1/2)i±(√((5i)/2))=  =−(1/2)±((√5)/2)+((1/2)±((√5)/2))i  z_1 =−((1+(√5))/2)+((1−(√5))/2)i  z_2 =−((1−(√5))/2)+((1+(√5))/2)i  (z+((1+(√5))/2)−((1−(√5))/2)i)(z+((1−(√5))/2)−((1+(√5))/2)i)=0
$${z}=−\frac{\mathrm{1}−\mathrm{i}}{\mathrm{2}}\pm\sqrt{\frac{\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{2}} }{\mathrm{4}}+\mathrm{3i}}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}\pm\sqrt{\frac{\mathrm{5i}}{\mathrm{2}}}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}+\left(\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\mathrm{i} \\ $$$${z}_{\mathrm{1}} =−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}+\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\mathrm{i} \\ $$$${z}_{\mathrm{2}} =−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}+\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\mathrm{i} \\ $$$$\left({z}+\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\mathrm{i}\right)\left({z}+\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\mathrm{i}\right)=\mathrm{0} \\ $$

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