Question Number 47291 by MJS last updated on 07/Nov/18
$$\mathrm{solve}\:\mathrm{for}\:{x}\in\mathbb{C}: \\ $$$$\mid{x}−\frac{\mathrm{3}}{\mathrm{4}}\mid×\mid{x}+\frac{\mathrm{5}}{\mathrm{4}}\mid=\mathrm{3} \\ $$
Answered by MrW3 last updated on 08/Nov/18
![let x+(1/4)=a+bi ∣x−(3/4)∣×∣x+(5/4)∣=3 (√((a−1)^2 +b^2 ))×(√((a+1)^2 +b^2 ))=3 [(a−1)^2 +b^2 ]×[(a+1)^2 +b^2 ]=9 (a^2 −1)^2 +b^2 [(a−1)^2 +(a+1)^2 ]+b^4 =9 b^4 +2(a^2 +1)b^2 +(a^2 +2)(a^2 −4)=0 b^2 =−(a^2 +1)±(√((a^2 +1)^2 −(a^2 +2)(a^2 −4))) b^2 =−(a^2 +1)±(√(4a^2 +9)) b^2 ≥0 −(a^2 +1)+(√(4a^2 +9))≥0 (√(4a^2 +9))≥a^2 +1 4a^2 +9≥a^4 +2a^2 +1 a^4 −2a^2 −8≤0 (a^2 +2)(a^2 −4)≤0 a^2 −4≤0 ⇒−2≤a≤2 ⇒b^2 =(√(4a^2 +9))−a^2 −1 ⇒b=±(√((√(4a^2 +9))−a^2 −1)) ⇒x=(a−(1/4))±i(√((√(4a^2 +9))−a^2 −1)) with −2≤a≤2](https://www.tinkutara.com/question/Q47336.png)
$${let}\:{x}+\frac{\mathrm{1}}{\mathrm{4}}={a}+{bi} \\ $$$$\mid{x}−\frac{\mathrm{3}}{\mathrm{4}}\mid×\mid{x}+\frac{\mathrm{5}}{\mathrm{4}}\mid=\mathrm{3} \\ $$$$\sqrt{\left({a}−\mathrm{1}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} }×\sqrt{\left({a}+\mathrm{1}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{3} \\ $$$$\left[\left({a}−\mathrm{1}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \right]×\left[\left({a}+\mathrm{1}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \right]=\mathrm{9} \\ $$$$\left({a}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \left[\left({a}−\mathrm{1}\right)^{\mathrm{2}} +\left({a}+\mathrm{1}\right)^{\mathrm{2}} \right]+{b}^{\mathrm{4}} =\mathrm{9} \\ $$$${b}^{\mathrm{4}} +\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right){b}^{\mathrm{2}} +\left({a}^{\mathrm{2}} +\mathrm{2}\right)\left({a}^{\mathrm{2}} −\mathrm{4}\right)=\mathrm{0} \\ $$$${b}^{\mathrm{2}} =−\left({a}^{\mathrm{2}} +\mathrm{1}\right)\pm\sqrt{\left({a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\left({a}^{\mathrm{2}} +\mathrm{2}\right)\left({a}^{\mathrm{2}} −\mathrm{4}\right)} \\ $$$${b}^{\mathrm{2}} =−\left({a}^{\mathrm{2}} +\mathrm{1}\right)\pm\sqrt{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{9}} \\ $$$${b}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$−\left({a}^{\mathrm{2}} +\mathrm{1}\right)+\sqrt{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{9}}\geqslant\mathrm{0} \\ $$$$\sqrt{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{9}}\geqslant{a}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} +\mathrm{9}\geqslant{a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1} \\ $$$${a}^{\mathrm{4}} −\mathrm{2}{a}^{\mathrm{2}} −\mathrm{8}\leqslant\mathrm{0} \\ $$$$\left({a}^{\mathrm{2}} +\mathrm{2}\right)\left({a}^{\mathrm{2}} −\mathrm{4}\right)\leqslant\mathrm{0} \\ $$$${a}^{\mathrm{2}} −\mathrm{4}\leqslant\mathrm{0} \\ $$$$\Rightarrow−\mathrm{2}\leqslant{a}\leqslant\mathrm{2} \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\sqrt{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{9}}−{a}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow{b}=\pm\sqrt{\sqrt{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{9}}−{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow{x}=\left({a}−\frac{\mathrm{1}}{\mathrm{4}}\right)\pm{i}\sqrt{\sqrt{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{9}}−{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$${with}\:−\mathrm{2}\leqslant{a}\leqslant\mathrm{2} \\ $$
Commented by MJS last updated on 09/Nov/18
$$\mathrm{thank}\:\mathrm{you}! \\ $$