The solution of the equation |z| - z = 1 + 2i | vedclass

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Question

(C) Given the equation $|z| - z = 1 + 2i$.Let $z = x + iy$,where $x, y \in \mathbb{R}$.Then $|x + iy| - (x + iy) = 1 + 2i$.$\sqrt{x^2 + y^2} - x - iy

Vedclass · Accepted Answer

$\frac{3}{2} - 2i$

Vedclass · Answer

(C) Given the equation $|z| - z = 1 + 2i$.Let $z = x + iy$,where $x, y \in \mathbb{R}$.Then $|x + iy| - (x + iy) = 1 + 2i$.$\sqrt{x^2 + y^2} - x - iy = 1 + 2i$.Equating the real and imaginary parts:$1$) Imaginary part: $-y = 2 \implies y = -2$.$2$) Real part: $\sqrt{x^2 + y^2} - x = 1$.Substitute $y = -2$ into the real part equation:$\sqrt{x^2 + (-2)^2} - x = 1$.$\sqrt{x^2 + 4} = 1 + x$.Squaring both sides:$x^2 + 4 = (1 + x)^2 = 1 + 2x + x^2$.$4 = 1 + 2x \implies 2x = 3 \implies x = \frac{3}{2}$.Thus,$z = \frac{3}{2} - 2i$.

The solution of the equation $|z| - z = 1 + 2i$ is

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