Question Number 194363 by sonukgindia last updated on 05/Jul/23
Commented by Frix last updated on 05/Jul/23
![There′s no solution. (√(z_1 +(√z_2 )))+(√(z_1 −(√z_2 )))=r Solving by 2 times [squaring (introduces false solutions) and transforming] leads to: z_1 =((r^4 +4z_2 +r^4 )/(4r^2 )) z_1 =a+bi∧z_2 =c+di with a, b, c, d ∈R a+bi=((4c+r^4 )/(4r^2 ))+(d/r^2 )i { ((a=((4c+r^4 )/(4r^2 )))),((b=(d/r^2 ))) :} We have r=2∧c=a+17 ⇒ a=7∧d=4b ⇒ f(b)=(√(7+bi+2(√(6+bi))))+(√(7+bi−2(√(6+bi)))) The real part of f(b) is ≥2(√6) The imaginary part of f(b) has only one zero at b=0 ⇒ f(0)=2(√6) is the only real point of f(b)](https://www.tinkutara.com/question/Q194385.png)
$$\mathrm{There}'\mathrm{s}\:\mathrm{no}\:\mathrm{solution}. \\ $$$$\sqrt{{z}_{\mathrm{1}} +\sqrt{{z}_{\mathrm{2}} }}+\sqrt{{z}_{\mathrm{1}} −\sqrt{{z}_{\mathrm{2}} }}={r} \\ $$$$\mathrm{Solving}\:\mathrm{by}\:\mathrm{2}\:\mathrm{times}\:\left[\mathrm{squaring}\:\left(\mathrm{introduces}\right.\right. \\ $$$$\left.\mathrm{f}\left.\mathrm{alse}\:\mathrm{solutions}\right)\:\mathrm{and}\:\mathrm{transforming}\right]\:\mathrm{leads}\:\mathrm{to}: \\ $$$${z}_{\mathrm{1}} =\frac{{r}^{\mathrm{4}} +\mathrm{4}{z}_{\mathrm{2}} +{r}^{\mathrm{4}} }{\mathrm{4}{r}^{\mathrm{2}} } \\ $$$${z}_{\mathrm{1}} ={a}+{b}\mathrm{i}\wedge{z}_{\mathrm{2}} ={c}+{d}\mathrm{i}\:\mathrm{with}\:{a},\:{b},\:{c},\:{d}\:\in\mathbb{R} \\ $$$${a}+{b}\mathrm{i}=\frac{\mathrm{4}{c}+{r}^{\mathrm{4}} }{\mathrm{4}{r}^{\mathrm{2}} }+\frac{{d}}{{r}^{\mathrm{2}} }\mathrm{i} \\ $$$$\begin{cases}{{a}=\frac{\mathrm{4}{c}+{r}^{\mathrm{4}} }{\mathrm{4}{r}^{\mathrm{2}} }}\\{{b}=\frac{{d}}{{r}^{\mathrm{2}} }}\end{cases} \\ $$$$\mathrm{We}\:\mathrm{have}\:{r}=\mathrm{2}\wedge{c}={a}+\mathrm{17} \\ $$$$\Rightarrow\:{a}=\mathrm{7}\wedge{d}=\mathrm{4}{b}\:\Rightarrow \\ $$$${f}\left({b}\right)=\sqrt{\mathrm{7}+{b}\mathrm{i}+\mathrm{2}\sqrt{\mathrm{6}+{b}\mathrm{i}}}+\sqrt{\mathrm{7}+{b}\mathrm{i}−\mathrm{2}\sqrt{\mathrm{6}+{b}\mathrm{i}}} \\ $$$$\mathrm{The}\:\mathrm{real}\:\mathrm{part}\:\mathrm{of}\:{f}\left({b}\right)\:\mathrm{is}\:\geqslant\mathrm{2}\sqrt{\mathrm{6}} \\ $$$$\mathrm{The}\:\mathrm{imaginary}\:\mathrm{part}\:\mathrm{of}\:{f}\left({b}\right)\:\mathrm{has}\:\mathrm{only}\:\mathrm{one} \\ $$$$\:\:\:\:\:\mathrm{zero}\:\mathrm{at}\:{b}=\mathrm{0} \\ $$$$\Rightarrow\:{f}\left(\mathrm{0}\right)=\mathrm{2}\sqrt{\mathrm{6}}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{real}\:\mathrm{point}\:\mathrm{of}\:{f}\left({b}\right) \\ $$
Answered by MacX last updated on 05/Jul/23
$$\sqrt{{z}+\sqrt{{z}+\mathrm{17}}}+\sqrt{{z}−\sqrt{{z}+\mathrm{17}}}=\mathrm{2}\: \\ $$$$\Rightarrow{z}+\sqrt{{z}+\mathrm{17}}+{z}−\sqrt{{z}+\mathrm{17}}+\mathrm{2}\sqrt{{z}^{\mathrm{2}} −\left({z}+\mathrm{17}\right)}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{2}{z}+\mathrm{2}\sqrt{{z}^{\mathrm{2}} −{z}−\mathrm{17}}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{2}\sqrt{{z}^{\mathrm{2}} −{z}−\mathrm{17}}=\mathrm{4}−\mathrm{2}{z}=\mathrm{2}\left(\mathrm{2}−{z}\right) \\ $$$$\Rightarrow{z}^{\mathrm{2}} −{z}−\mathrm{17}={z}^{\mathrm{2}} −\mathrm{4}{z}+\mathrm{4} \\ $$$$\Rightarrow−{z}−\mathrm{17}=−\mathrm{4}{z}+\mathrm{4} \\ $$$$\Rightarrow\mathrm{3}{z}=\mathrm{21} \\ $$$$\Rightarrow{z}=\mathrm{7} \\ $$$${S}_{\mathbb{C}} =\left\{\mathrm{7}\right\} \\ $$
Commented by Frix last updated on 05/Jul/23
$$\sqrt{\mathrm{7}+\sqrt{\mathrm{7}+\mathrm{17}}}+\sqrt{\mathrm{7}−\sqrt{\mathrm{7}+\mathrm{17}}}=\sqrt{\mathrm{7}+\sqrt{\mathrm{24}}}+\sqrt{\mathrm{7}−\sqrt{\mathrm{24}}}= \\ $$$$=\left(\mathrm{1}+\sqrt{\mathrm{6}}\right)+\left(−\mathrm{1}+\sqrt{\mathrm{6}}\right)=\mathrm{2}\sqrt{\mathrm{6}}\neq\mathrm{2} \\ $$