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

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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,

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

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