Mathematics - Complex Number Question with Solution | TestHub

Given,

${\left|z\right|}^3+2z^2+4\overline{z}-8=0 .......\left(1\right)$

Now on taking conjugate both side we get,

${\left|z\right|}^3+2{\overline{z}}^2+4z-8=0 ........\left(2\right)$

$\left\{\text{Note} \left|\overline{z}\right|=\left|z\right|\right\}$

Now subtracting both equations we get,

$2\left(z^2-{\overline{z}}^2\right)+4\left(\overline{z}-z\right)=0$ 

$\Rightarrow \left(z-\overline{z}\right)\left[2\left(z+\overline{z}\right)-4\right]=0$

$\because z=\overline{z}$ (not possible) or $4x=4\Rightarrow x=1$ $\left\{\text{as} z+\overline{z}=2x\right\}$

So, $z=1+\lambda i$

$\Rightarrow \left|z\right|=\sqrt{1+{\lambda }^2} \& \overline{z}=1-\lambda i$

Now putting the value of $z, \left|z\right| \& \overline{z}$ in given equation we get, 

${\left(1+{\lambda }^2\right)}^{3/2}+2\left(1-{\lambda }^2+2\lambda i\right)+4\left(1-\lambda i\right)-8=0$

$\Rightarrow {\left(1+{\lambda }^2\right)}^{3/2}+2\left(1-{\lambda }^2\right)=4$

$\Rightarrow {\left(1+{\lambda }^2\right)}^{3/2}=2\left(1+{\lambda }^2\right)$

$\Rightarrow \left(1+{\lambda }^2\right)\left[\sqrt{1+{\lambda }^2}-2\right]=0$

$\Rightarrow {\lambda }^2=3$

Now solving,

$\left(\mathrm{P}\right)$ ${\left|z\right|}^2=1+{\lambda }^2=1+3=4$

$\left(\mathrm{Q}\right)$ ${\left|z-\overline{z}\right|}^2={\left|1+\lambda i-\left(1-\lambda i\right)\right|}^2={\left|2\lambda i\right|}^2=4{\lambda }^2=12$

$\left(\mathrm{R}\right)$ ${\left|z\right|}^2+{\left|z+\overline{z}\right|}^2=4+{\left|\left(1+\lambda i\right)+\left(1-\lambda i\right)\right|}^2=4+4=8$

$\left(\mathrm{S}\right)$ ${\left|z+1\right|}^2={\left|1+\lambda i+1\right|}^2=4+{\lambda }^2=4+3=7$

$\therefore \mathrm{P}\to \left(2\right), \mathrm{Q}\to \left(1\right), \mathrm{R}\to \left(3\right), \mathrm{S}\to \left(5\right)$