If z is a complex number such that z + 1 + i = |z|, then what statements about z are correct?

Steps to Solve

1. Rewrite the given equation
   We begin with the equation involving the complex number $z$: $$ z + 1 + i = |z| $$

2. Isolate the complex number z
   We can rewrite the equation to isolate $z$: $$ z = |z| - 1 - i $$

3. Calculate the modulus of z
   The modulus ($|z|$) of a complex number $z = x + yi$ (where $x$ is the real part and $y$ is the imaginary part) is given by: $$ |z| = \sqrt{x^2 + y^2} $$

From our expression for $z$, we have: $$ z = (|z| - 1) + (-1)i $$ Thus,

• Real part: $x = |z| - 1$
• Imaginary part: $y = -1$

Calculate $|z|$ from the real and imaginary parts: $$ |z| = \sqrt{(|z| - 1)^2 + (-1)^2} $$

4. Square both sides
   Square both sides of the modulus equation to eliminate the square root: $$ |z|^2 = (|z| - 1)^2 + 1 $$

5. Expand and simplify
   Expanding the right side: $$ |z|^2 = (|z|^2 - 2|z| + 1) + 1 $$
   This simplifies to: $$ |z|^2 = |z|^2 - 2|z| + 2 $$

6. Rearrange the equation
   Subtract $|z|^2$ from both sides: $$ 0 = -2|z| + 2 $$

7. Solve for |z|
   Isolate $|z|$: $$ 2|z| = 2 $$ $$ |z| = 1 $$

8. Evaluate the provided statements
   A. $|z| = 1$: This is true.

To examine the other statements:

• B. $z = \overline{z}$ implies $z$ is real; this could be possible since $z$ has a modulus of 1 but needs confirmation through argument comparison.
• C. $z = -\overline{z}$ suggests $z$ is purely imaginary, which conflicts with the modulus.
• D. $\arg(z) = \frac{\pi}{2}$ would imply $z$ is purely imaginary, thus cannot satisfy $|z|=1$.