Determine the complex number z that satisfies the equation…

Determine the complex number $z$ that satisfies the equation $|z|+z=2+i$.

Step 1 · Set up in Cartesian form

Let $z=x+iy$. Then:

$$ |z|+z = \sqrt{x^2+y^2} + x + iy = \left(\sqrt{x^2+y^2}+x\right) + iy $$

Step 2 · Equate real and imaginary parts

Setting this equal to $2+i$:

$$ \text{Imaginary: } y = 1 \qquad\qquad \text{Real: } \sqrt{x^2+y^2}+x = 2 $$

Step 3 · Substitute $y=1$

$$ \sqrt{x^2+1}+x = 2 \quad\Longrightarrow\quad \sqrt{x^2+1} = 2-x $$

Step 4 · Solve for $x$

For the square root (nonnegative) to equal $2-x$, we need $2-x\ge0$, i.e. $x\le2$. Squaring both sides:

$$ x^2+1 = (2-x)^2 = 4-4x+x^2 \quad\Longrightarrow\quad 1 = 4-4x \quad\Longrightarrow\quad x = \frac34 $$

Step 5 · Check the sign condition

$x=\dfrac34 \le 2$ ✓, and $2-x=\dfrac54>0$ ✓, so squaring introduced no extraneous solution.

Step 6 · Write the answer

$$ \boxed{z = \frac34+i} $$

Step 7 · Verify

$$ |z| = \sqrt{\left(\frac34\right)^2+1^2} = \sqrt{\frac9{16}+\frac{16}{16}} = \sqrt{\frac{25}{16}} = \frac54 $$

$$ |z|+z = \frac54+\frac34+i = \frac84+i = 2+i \quad\checkmark $$

Step 8 · Uniqueness

The imaginary part forces $y=1$ with no alternative. For the real part, $f(x)=\sqrt{x^2+1}+x$ is strictly increasing (both terms increase as $x$ grows), so the equation $f(x)=2$ has at most one solution — hence $x=\dfrac34$ is the only one, and $z=\dfrac34+i$ is the unique solution overall.

Step 9 · Geometric interpretation

Rewriting the equation as $|z| = (2+i)-z$, and taking moduli of both sides gives $|z| = |(2+i)-z|$ — meaning $z$ is equidistant from the origin and from the point $2+i$, i.e. $z$ lies on the perpendicular bisector of the segment from $0$ to $2+i$. Checking: $|z|=\dfrac54$ and $|(2+i)-z| = \left|\dfrac54\right| = \dfrac54$. ✓