Question-196893

Question Number 196893 by hardmath last updated on 02/Sep/23

Answered by Rasheed.Sindhi last updated on 02/Sep/23

x=(√((−1+i)+(√((−1+i)+(√(...)))))) x=(√((−1+i)+x)) x^2 = (−1+i)+x x^2 −x+1−i=0 x=((1±(√(1−4(1−i))))/2) x=((1±(√(−3+4i)))/2) =((1±(1+2i))/2) =((2+2i)/2) , ((−2i)/2) =1+i , −i

$${x}=\sqrt{\left(−\mathrm{1}+{i}\right)+\sqrt{\left(−\mathrm{1}+{i}\right)+\sqrt{…}}}\: \\ $$$${x}=\sqrt{\left(−\mathrm{1}+{i}\right)+{x}}\: \\ $$$${x}^{\mathrm{2}} =\:\left(−\mathrm{1}+{i}\right)+{x} \\ $$$${x}^{\mathrm{2}} −{x}+\mathrm{1}−{i}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}\left(\mathrm{1}−{i}\right)}}{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{1}\pm\sqrt{−\mathrm{3}+\mathrm{4}{i}}}{\mathrm{2}} \\ $$$$\:\:\:\:=\frac{\mathrm{1}\pm\left(\mathrm{1}+\mathrm{2}{i}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:=\frac{\mathrm{2}+\mathrm{2}{i}}{\mathrm{2}}\:,\:\frac{−\mathrm{2}{i}}{\mathrm{2}} \\ $$$$\:\:\:\:\:=\mathrm{1}+{i}\:,\:−{i} \\ $$

Commented by BaliramKumar last updated on 02/Sep/23