Question

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

Answer 1

`z=+-(1/2){sqrt(sqrt19-4)+isqrt(sqrt19+4)/sqrt3}.`

Explanation:

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

Also, `z^2=(x+iy)^2=x^2+2ixy+i^2y^2=x^2-y^2+2ixy.`

With these, the eqn. becomes,

`x^2+y^2-2x^2+2y^2-4ixy=2-i, i.e.,`

`3y^2-x^2-i(4xy)=2-i.`

Comparing the Real & Imaginary Parts, we get,

`3y^2-x^2=2... ...(1), and, 4xy=1 rArr y=1/(4x)...(2).`

`:. 3(1/(4x))^2-x^2=2 rArr 3-16x^4=32x^2, or,`

`16x^4+32x^2=3 :. x^4+2x^2=3/16.`

Completing the Square, `x^4+2x^2+1=3/16+1=19/16.`

`:. (x^2+1)^2=19/16 rArr x^2+1=+-sqrt19/4.`

But, `x^2+1>0 rArr x^2+1=+sqrt19/4 rArr x^2=(sqrt19-4)/4.`

`:. x=+-sqrt(sqrt19-4)/2`

`(1) rArr 3y^2=x^2+2=(x^2+1)+1=(sqrt19/4)+1=(sqrt19+4)/4.`

`:. (sqrt3)y=+-sqrt(sqrt19+4)/2,` giving,

`y=+-sqrt(sqrt19+4)/(2sqrt3).`

`:." The Soln. is, "z=x+iy=+-(1/2){sqrt(sqrt19-4)+isqrt(sqrt19+4)/sqrt3}.`

I leave it to the Reader to verify this soln.

Enjoy Maths.!

Answer 2

See below.

Explanation:

Calling `z=(x+iy)` and `bar z = (x-iy)` we have

`z bar z-2z^2=2-i`

or

`x^2-3y^2+2-i(4xy-1)=0+i0`

Solving now

`{(x^2-3y^2+2=0),(4xy-1=0):}`

or almost equivalently

`{(x^2-3y^2+2=0),(16x^2y^2=1):}`

Solving for `x^2, y^2` we have

`((x^2,y^2),(1/4 (-4 - sqrt[19]), 1/12 (4 - sqrt[19])),(1/4 (-4 + sqrt[19]), 1/12 (4 + sqrt[19])))`

or

`z_(1,2)=i/2( pmsqrt(sqrt(19)+4)pmisqrt((sqrt(19)-4)/3))`
`z_(3,4)=1/2(pmsqrt(sqrt(19)-4)pm isqrt((sqrt(19)+4)/3))`

giving a total of four solutions.