How to determine complex numbers that satisfy relation:#abs(z)^2-2iz+2a(1+i)=0;a>0#?

Question

2363 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-03T18:34:55

Write #z=x+"i"y#. Simplify. Equate real and imaginary parts, solve.

#z=x+"i"y#

implies that,

#abs(z)^2=x^2+y^2#,

#2"i"z=2"i"x-2y#.

Substituting,

#x^2+y^2-2"i"x+2y+2a+2a"i"=0#,

#(x^2+y^2+2y+2a)+"i"(2a-2x)=0#.

The imaginary and real parts of the RHS must equal the imaginary and real parts of the LHS as #x# and #y# are both real.

Then,

#2a-2x=0# and #x^2+y^2-2y+2a=0#.

Solving the first gives #a=x#. Substituting this into the second gives,

#y^2+2y+a^2+2a=0#,
#(y+1)^2+a^2+2a-1=0#,
#(y+1)^2+(a+1)^2-2=0#,
#y=-1 \pm sqrt(2-(a+1)^2)#.

Then the solutions for #z# are,

#z=a+"i"(-1+ sqrt(2-(a+1)^2))#,
#z=a+"i"(-1- sqrt(2-(a+1)^2))#.