# How to solve this equation for z, where z is a complex number: (|z|)^(2)-2(z^(2))=2-i?

Mar 8, 2017

$z = \pm \left(\frac{1}{2}\right) \left\{\sqrt{\sqrt{19} - 4} + i \frac{\sqrt{\sqrt{19} + 4}}{\sqrt{3}}\right\} .$

#### Explanation:

Let, z=x+iy; x,y in RR rArr |z|^2=x^2+y^2.

Also, ${z}^{2} = {\left(x + i y\right)}^{2} = {x}^{2} + 2 i x y + {i}^{2} {y}^{2} = {x}^{2} - {y}^{2} + 2 i x y .$

With these, the eqn. becomes,

${x}^{2} + {y}^{2} - 2 {x}^{2} + 2 {y}^{2} - 4 i x y = 2 - i , i . e . ,$

$3 {y}^{2} - {x}^{2} - i \left(4 x y\right) = 2 - i .$

Comparing the Real & Imaginary Parts, we get,

$3 {y}^{2} - {x}^{2} = 2. . . \ldots \left(1\right) , \mathmr{and} , 4 x y = 1 \Rightarrow y = \frac{1}{4 x} \ldots \left(2\right) .$

$\therefore 3 {\left(\frac{1}{4 x}\right)}^{2} - {x}^{2} = 2 \Rightarrow 3 - 16 {x}^{4} = 32 {x}^{2} , \mathmr{and} ,$

$16 {x}^{4} + 32 {x}^{2} = 3 \therefore {x}^{4} + 2 {x}^{2} = \frac{3}{16.}$

Completing the Square, ${x}^{4} + 2 {x}^{2} + 1 = \frac{3}{16} + 1 = \frac{19}{16.}$

$\therefore {\left({x}^{2} + 1\right)}^{2} = \frac{19}{16} \Rightarrow {x}^{2} + 1 = \pm \frac{\sqrt{19}}{4.}$

But, ${x}^{2} + 1 > 0 \Rightarrow {x}^{2} + 1 = + \frac{\sqrt{19}}{4} \Rightarrow {x}^{2} = \frac{\sqrt{19} - 4}{4.}$

$\therefore x = \pm \frac{\sqrt{\sqrt{19} - 4}}{2}$

$\left(1\right) \Rightarrow 3 {y}^{2} = {x}^{2} + 2 = \left({x}^{2} + 1\right) + 1 = \left(\frac{\sqrt{19}}{4}\right) + 1 = \frac{\sqrt{19} + 4}{4.}$

$\therefore \left(\sqrt{3}\right) y = \pm \frac{\sqrt{\sqrt{19} + 4}}{2} ,$ giving,

$y = \pm \frac{\sqrt{\sqrt{19} + 4}}{2 \sqrt{3}} .$

$\therefore \text{ The Soln. is, } z = x + i y = \pm \left(\frac{1}{2}\right) \left\{\sqrt{\sqrt{19} - 4} + i \frac{\sqrt{\sqrt{19} + 4}}{\sqrt{3}}\right\} .$

I leave it to the Reader to verify this soln.

Enjoy Maths.!

Mar 8, 2017

See below.

#### Explanation:

Calling $z = \left(x + i y\right)$ and $\overline{z} = \left(x - i y\right)$ we have

$z \overline{z} - 2 {z}^{2} = 2 - i$

or

${x}^{2} - 3 {y}^{2} + 2 - i \left(4 x y - 1\right) = 0 + i 0$

Solving now

$\left\{\begin{matrix}{x}^{2} - 3 {y}^{2} + 2 = 0 \\ 4 x y - 1 = 0\end{matrix}\right.$

or almost equivalently

$\left\{\begin{matrix}{x}^{2} - 3 {y}^{2} + 2 = 0 \\ 16 {x}^{2} {y}^{2} = 1\end{matrix}\right.$

Solving for ${x}^{2} , {y}^{2}$ we have

$\left(\begin{matrix}{x}^{2} & {y}^{2} \\ \frac{1}{4} \left(- 4 - \sqrt{19}\right) & \frac{1}{12} \left(4 - \sqrt{19}\right) \\ \frac{1}{4} \left(- 4 + \sqrt{19}\right) & \frac{1}{12} \left(4 + \sqrt{19}\right)\end{matrix}\right)$

or

${z}_{1 , 2} = \frac{i}{2} \left(\pm \sqrt{\sqrt{19} + 4} \pm i \sqrt{\frac{\sqrt{19} - 4}{3}}\right)$
${z}_{3 , 4} = \frac{1}{2} \left(\pm \sqrt{\sqrt{19} - 4} \pm i \sqrt{\frac{\sqrt{19} + 4}{3}}\right)$

giving a total of four solutions.