The no of complex numbers satisfying $| z | = z + 1 + 2 i$ .

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Solution

Let $z = a + i b$
$| z | = \sqrt{a^{2} + b^{2}}$
$⟹ \sqrt{a^{2} + b^{2}} = a + i b + 1 + 2 i$
$⟹ \sqrt{a^{2} + b^{2}} = a + 1 + i (b + 2)$
$⟹ b + 2 = 0 ⟹ b = - 2$
and $\sqrt{a^{2} + b^{2}} = a + 1$
$⟹ a^{2} + 4 = a^{2} + 1 + 2 a ⟹ 2 a = 3 ⟹ a = 3 / 2$
So, $z = \frac{3}{2} - 2 i$

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