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

z = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} a (- 1 \pm \sqrt{1 - a^{2} - 2 a}) i, & 0 < a < \sqrt{2} - 1 (\sqrt{2} - 1) - i, & a = \sqrt{2} - 1 no solution, & a > \sqrt{2} - 1 \end{matrix}

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z = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} a (1 \pm \sqrt{1 - a^{2} - 2 a}) i, & 0 < a < \sqrt{2} - 1 (\sqrt{2} - 1) - i, & a = \sqrt{2} - 1 no solution, & a > \sqrt{2} - 1 \end{matrix}

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z = {\begin{matrix} a (- 1 \pm \sqrt{1 - a^{2} - 2 a}) i, & 0 < a < \sqrt{2} - 1 (\sqrt{2} - 1) - i, & a \geq \sqrt{2} - 1 \end{matrix}

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z = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} a (- 1 \pm \sqrt{1 - a^{2} - 2 a}) i, & 0 < a < \sqrt{2} + 1 \sqrt{2} - 1) - i, & a = \sqrt{2} + 1 no solution, & a > \sqrt{2} + 1 \end{matrix}

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Solution

The correct option is A $z = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} a (- 1 \pm \sqrt{1 - a^{2} - 2 a}) i, & 0 < a < \sqrt{2} - 1 (\sqrt{2} - 1) - i, & a = \sqrt{2} - 1 no solution, & a > \sqrt{2} - 1 \end{matrix}$
Put $z = x + i y$ .
Then the given equation reduces to
$x^{2} + y^{2} - 2 i (x + i y) + 2 a (1 + i) = 0$
Or $(x^{2} + y^{2} + 2 y + 2 a) + 2 i (a - x) = 0$
Equating the real and imaginary part to zero, we get
$x^{2} + y^{2} + 2 y + 2 a = 0, 2 a - 2 x = 0$
This gves for $x = a$ and for y we have
$a^{2} + y^{2} + 2 y + 2 a = 0$ or $y^{2} + 2 y + (a^{2} + 2 a) = 0$ ...(1)
Since we seek real value of $y$ , the discriminant of the equation (1)
must be non-negative, that is
$△ = 4 - 4 (a^{2} + 2 a) \geq 0 \Rightarrow 1 - a^{2} - 2 a \geq 0$
For the real value of $a$ , we get
$y = \frac{- 2 \pm \sqrt{4 (1 - a^{2} - 2 a)}}{2} = - 1 \pm \sqrt{1 - a^{2} - 2 a}$
Hence for $△ \geq 0$ , the original equation has two roots
$z_{1} = a + (- 1 + \sqrt{1 - a^{2} + 2 a}) i z_{2} = a + (- 1 - \sqrt{1 - a^{2} + 2 a}) i$
For $△ = 0$ , we have $z_{1} = z_{2}$ that is,
there is only one solution in this case.
For $△ < 0$ the equation has no roots
It remain to indicate the range of $a$ over which there are solution.
We are given $a \geq 0$ and in addition to this we found that $a$
must satisfy the inequality.
$1 - a^{2} - 2 a \geq 0 \Rightarrow a^{2} + 2 a - 1 \leq 0 \Rightarrow {(a + 1)}^{2} - 2 \leq 0 {(a - 1)}^{2} \leq 2 \Rightarrow - \sqrt{2} \leq a + 1 \leq \sqrt{2} \Rightarrow - 1 - \sqrt{2} \leq a \leq - 1 + \sqrt{2}$
Now choosing the number $a \geq 0$ from this interval we obtain
$0 \leq a \leq - 1 + \sqrt{2}$
Thus
$z = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} a (- 1 \pm \sqrt{1 - a^{2} - 2 a}) i, & 0 < a < \sqrt{2} - 1 (\sqrt{2} - 1) - i, & a = \sqrt{2} - 1 no solution, & a > \sqrt{2} - 1 \end{matrix}$

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