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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

If complex numbers $(x -2y) + i(3x -y)$ and $(2x -y) + i(x -y + 6)$ are conjugates of each other, then $|x + iy|$ is $(x,y \in R)$

The sum of amplitude of $z$ and another complex number is $\pi $. The other complex number can be written

If $\alpha $ and $\beta $ are different complex numbers with $|\beta | = 1$, then $\left| {\frac{{\beta - \alpha }}{{1 - \overline \alpha \beta }}} \right|$ is equal to

If ${z_1}$ and ${z_2}$ are two complex numbers, then $|{z_1} - {z_2}|$ is

Conjugate of $1 + i$ is