Animated Solution for Mathematics - Complex Numbers: If $|z-4|<|z-2|$, its solution is given by

The problem asks us to solve the inequality $|z-4|<|z-2|$ in the complex plane. In the Argand plane, $|z-a|$ represents the distance between the complex number $z$ and the point $a$.

Therefore, the inequality $|z-4|<|z-2|$ asks us to identify all points $z$ that are strictly closer to the point $4$ than they are to the point $2$.

Geometrically, the points equidistant from $2$ and $4$ lie on the perpendicular bisector of the segment connecting them, which is the vertical line $x=3$. Points to the right of this line are closer to $4$, while points to the left are closer to $2$.

To solve this rigorously, we substitute $z=x+iy$, where $x=\text{Re}(z)$ and $y=\text{Im}(z)$. The inequality becomes:

Using the definition of the modulus $|a+ib|=\sqrt{a^2+b^2}$, we rewrite the expression as:

Since both sides represent distances and are non-negative, we can safely square both sides without altering the inequality:

Expanding the squared terms on both sides, we obtain:

Notice that the $x^2$ and $y^2$ terms appear on both sides of the inequality. We can cancel these terms out, which simplifies the expression significantly:

Now, we isolate $x$ by rearranging the terms:

Dividing both sides by $4$, we arrive at the final result:

We have determined that the real part of $z$ must be greater than $3$. This confirms our initial geometric intuition: the region satisfying the inequality is the half-plane defined by $\text{Re}(z)>3$.

By combining geometric visualization with algebraic rigor, we have mastered the problem. This approach ensures that you can handle similar problems in the JEE Advanced with confidence and clarity.