For every real value of $a > 0$ , determine the complex numbers which will satisfy the equation $| z |^{2} - 2 i z + 2 a (1 + i) = 0$ .

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Solution

Write $z = x + i y$ equal real and imaginary parth, solve
$| z |^{2} = x^{2} + y^{2}$
$2 i = 2 i x - 2 y$
Substituting
$x^{2} + y^{2} - 2 i x + 2 y + 2 a + 2 a i = 0$
$(2^{2} + y^{2} - 2 y + 2 a) + i (2 a - 2 x) = 0$
The real and imaginary parts of the $R H S$ most equal the real and imaginary parts of the $L H S$ as $x$ and $y$ both real.
Then
$2 a - 2 x = 0$ and $x^{2} + y^{2} - 2 y + 2 a = 0$
Solving the first givan $a = 0$
Substituting thes into the second given
$y^{2} + 2 y + a^{2} + 2 a = 0$
$(y + 1)^{2} + a^{2} + 2 a - 1 = 0$
$(y + 1)^{2} + (a + 1)^{2} - 2 = 0$
$y = - 1 \pm \sqrt{2 - (a + 1)^{2}}$
Then the solution for $z$ are
$z = a + i (- 1 + \sqrt{2 - (a + 1)^{2}})$
$z = a + i (- 1 - \sqrt{2 - (a + 1)^{2}})$

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