Animated Solution for Mathematics - Complex Numbers: If $z$ be a complex number satisfying $|Re(z)|+|Im(z)|=4$, then $|z|$ cannot be:

The given condition is $|Re(z)|+|Im(z)|=4$. By representing $z$ as $x+iy$, the condition transforms into the equation:

This equation does not represent a circle, but rather a square rotated by $45^{\circ }$. In the first quadrant, where $x,y>0$, the equation simplifies to $x+y=4$.

Due to the symmetry across all four quadrants, this shape forms a perfect square with vertices at $(4,0)$, $(0,4)$, $(-4,0)$, and $(0,-4)$.

The modulus $|z|$ represents the distance of a point $z$ from the origin $(0,0)$. To find the range of $|z|$, we identify the minimum and maximum distances from the origin to the boundary of the square.

The maximum distance occurs at the vertices of the square. The distance from the origin to any vertex (e.g., $(4,0)$) is:

The minimum distance occurs at the midpoints of the sides of the square. Using the perpendicular distance formula from the origin $(0,0)$ to the line $x+y=4$:

Thus, the modulus $|z|$ must lie within the closed interval:

The range $[\sqrt{8},\sqrt{16}]$ defines the set of all possible distances from the origin to any point on the square. Any value of $|z|$ falling outside this interval is impossible.

Comparing the values: 1. $\sqrt{8}\approx 2.828$ 2. $\sqrt{7}\approx 2.645$

Since $\sqrt{7}<\sqrt{8}$, the value $\sqrt{7}$ lies strictly inside the square, meaning it does not satisfy the condition $|Re(z)|+|Im(z)|=4$. Therefore, $\sqrt{7}$ is not a possible value for $|z|$.

This problem demonstrates the power of geometric intuition in the Argand plane. By visualizing the algebraic condition as a geometric locus, we avoid tedious calculations and identify the "intruder" value immediately.

Always remember these fundamental geometric interpretations for JEE Advanced: $|Re(z)|+|Im(z)|=k$ represents a square. $|z|=k$ represents a circle centered at the origin. $|z-z_0|=k$ represents a circle* centered at $z_0$.