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

2 Answers
Mar 8, 2017

# 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.!

Mar 8, 2017

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.