Indian Navy SSR Mathematics — practice by topic

$\tan\theta = \dfrac{y}{x} = 1$, and $z$ is in the first quadrant, so $\theta = \dfrac{\pi}{4}$.

**Geometric reading.** $|z - 2|$ is the distance from the point $z$ to $(2, 0)$. $|z + 2|$ is the distance to $(-2, 0)$. Equating them means $z$ is equidistant from the two points. **Locus.** Set of points equidistant from $(2, 0)$ and $(-2, 0)$ is the **perpendicular bisector** of the segment joining them — the $y$-axis, i.e. $\mathrm{Re}(z) = 0$. **Algebraic confirmation.** Let $z = x + iy$. Squaring: $(x-2)^2 + y^2 = (x+2)^2 + y^2 \Rightarrow -4x = 4x \Rightarrow x = 0$. **Why option C is tempting.** A symmetric-looking radius-2 hint, but $|z - 2|$ is *not* the same as $|z| - 2$ it's a distance to a *point*, not a centred circle equation.

3D distance formula extends Pythagoras into three dimensions: the distance is the square root of the sum of squared coordinate differences. Derivation: project the segment $PQ$ onto each axis to get the leg lengths $|x_2 - x_1|$, etc. Apply Pythagoras twice. Option B is the **Manhattan** (taxicab) distance, useful in some lattice problems but not the Euclidean answer.