Let z be a complex number such that |z|+z=2+i | vedclass

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

Question

(B) Given $|z|+z=2+i$. Let $z=x+iy$,where $x, y \in \mathbb{R}$. Then $|z|=\sqrt{x^2+y^2}$. Substituting these into the equation: $\sqrt{x^2+y^2}+x+iy

Vedclass · Accepted Answer

$\frac{5}{4}$

Vedclass · Answer

(B) Given $|z|+z=2+i$. Let $z=x+iy$,where $x, y \in \mathbb{R}$. Then $|z|=\sqrt{x^2+y^2}$. Substituting these into the equation: $\sqrt{x^2+y^2}+x+iy=2+i$. Equating real and imaginary parts,we get: $y=1$ and $\sqrt{x^2+y^2}+x=2$. Substituting $y=1$ into the second equation: $\sqrt{x^2+1}=2-x$. Squaring both sides: $x^2+1=(2-x)^2 = 4-4x+x^2$. $1=4-4x$ $\Rightarrow 4x=3$ $\Rightarrow x=\frac{3}{4}$. Thus,$z=\frac{3}{4}+i$. Therefore,$|z|=\sqrt{(\frac{3}{4})^2+1^2} = \sqrt{\frac{9}{16}+1} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

Let $z$ be a complex number such that $|z|+z=2+i$,where $i=\sqrt{-1}$. Then $|z|$ is equal to:

Similar Questions

If $\frac{(p + i)^2}{2p - i} = \mu + i\lambda$,then $\mu^2 + \lambda^2$ is equal to

Sum of the moduli of the complex roots of the equation $(x^2+\frac{1}{x^2})-5(x+\frac{1}{x})+6=0$ is

If the point $z=(1+i)(1+2i)(1+3i) \ldots (1+10i)$ lies on a circle with centre at the origin and radius $r$,then $r^2$ is equal to

The modulus of the square root of the conjugate of $-7+24 \sqrt{-1}$ is ....

If $\alpha_1, \alpha_2, \alpha_3$ respectively denote the moduli of the complex numbers $-i, \frac{1}{3}(1+i)$ and $-1+i$,then their increasing order is

Vedclass Test Series

Exam Paper Generator

Online Exam Module