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

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(c)$|z| - z = 1 + 2i$
Let $z = x + iy$, therefore $|x + iy| - (x + iy) = 1 + 2i$
Equating real and imaginary parts, we get
$\sqrt {{x^2} + {y^2}} -

Vedclass · Accepted Answer

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

Vedclass · Answer

(c)$|z| - z = 1 + 2i$
Let $z = x + iy$, therefore $|x + iy| - (x + iy) = 1 + 2i$
Equating real and imaginary parts, we get
$\sqrt {{x^2} + {y^2}} - x = 1$and $y = - 2$==>$x = \frac{3}{2}$
Hence complex number $z = \frac{3}{2} - 2i$.
Trick : Since $\left| {\frac{3}{2} - 2i} \right| - \left( {\frac{3}{2} - 2i} \right)$
$ = \sqrt {\frac{9}{4} + 4} - \frac{3}{2} + 2i = \frac{5}{2} - \frac{3}{2} + 2i = 1 + 2i$

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

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