The inequality |z-4|

Q: Wait, why do we reject the negative root here?

Actually, we do not have a negative root to reject! When we square both sides of the inequality $|z - 4| < |z - 2|$, we are squaring two non-negative real numbers (distances). Since the square function is monotonically increasing for positive values, the inequality sign remains unchanged. No roots are involved in the final simplification.

Q: Why is the boundary a vertical line?

Great observation! The boundary is the perpendicular bisector of the segment joining $z_1 = 2$ and $z_2 = 4$. Since both points lie on the real axis, the segment is horizontal. The perpendicular bisector of a horizontal segment must be a vertical line. In this case, it is the line $x = 3$.

Q: Why is the answer 'none of these'?

The inequality simplifies to $x > 3$, which means the real part of $z$ must be strictly greater than $3$. Looking at the options provided, none of them ($Re(z) \ge 0$, $Re(z) 0$) are equivalent to $Re(z) > 3$. Therefore, 'none of these' is the correct choice.

Question

The inequality $|z-4| < |z-2|$ represents the region given by

Options:

(A) Re(z) \ge 0

(B) Re(z) < 0

(C) Re(z) > 0

(D) none of these

Sigma Percentile · Accepted Answer

### The Geometry of Complex Numbers: A Journey into the Argand Plane Imagine you are standing on the complex plane, a vast, two-dimensional landscape where every point is a complex number $z = x + iy$. Today, we are exploring a beautiful inequality: $|z - 4| < |z - 2|$. At first glance, this might look like a dry algebraic exercise, but it is actually a story about distance and territory. ### The Concept of Distance In the complex plane, the modulus $|z - z_0|$ is not just a symbol; it is the physical distance between your variable point $z$ and a fixed landmark $z_0$. So, when you see $|z - 4| < |z - 2|$, you are essentially asking: "Where can I stand such that I am closer to the point $4$ than I am to the point $2$?" It is a tug-of-war between two landmarks on the real axis. ### Finding the Boundary Before we solve for the entire region, let us find the "no-man's-land"—the boundary where you are exactly equidistant from both. This is the perpendicular bisector of the segment connecting $2$ and $4$. The midpoint is $3$, and since the segment is horizontal, the perpendicular bisector is the vertical line $x = 3$. Any point on this line is perfectly balanced between the two landmarks. ### The Algebraic Proof Now, let us bring in the rigor. We substitute $z = x + iy$ into our inequality: $$| (x - 4) + iy | < | (x - 2) + iy |$$ Using the definition of the modulus, this becomes: $$\sqrt{(x - 4)^2 + y^2} < \sqrt{(x - 2)^2 + y^2}$$ Since both sides represent distances, they are inherently positive. We can safely square both sides without flipping the inequality: $$(x - 4)^2 + y^2 < (x - 2)^2 + y^2$$ Expanding these squares, we get: $$x^2 - 8x + 16 + y^2 < x^2 - 4x + 4 + y^2$$ The $x^2$ and $y^2$ terms cancel out beautifully, leaving us with: $$-8x + 16 < -4x + 4$$ Rearranging this, we find $12 < 4x$, or simply **$x > 3$**. ### The Final Revelation Our algebraic result, $x > 3$, confirms our geometric intuition. The region is the entire half-plane to the right of the vertical line $x = 3$. When we look at the options provided—$Re(z) \ge 0$, $Re(z) < 0$, $Re(z) > 0$—none of them capture the specific requirement that the real part must be greater than $3$. Thus, we confidently select **"none of these"**. You have just mastered the art of visualizing complex inequalities!

Animated Solution for Mathematics - Complex Numbers: The inequality $∣ z - 4∣ < ∣ z - 2∣$ represents the region given by

Select Answer:

Visualized Solution

The Given Inequality

Geometric Interpretation

Locating the Fixed Points

Visualizing Distances

The Boundary Condition

The Line $x = 3$

Algebraic Substitution

Grouping Terms

Modulus Formula

Squaring Both Sides

Expanding the Squares

Canceling Common Terms

Final Inequality for $x$

The Region $R e (z) > 3$

Checking the Options

The Sigma Insight: Geometrical Applications of Complex Numbers

The Geometry of Complex Numbers

The Concept of Distance

Finding the Boundary

The Algebraic Proof

The Final Revelation

Similar Questions

If $∣ z - 4∣ < ∣ z - 2∣$ , its solution is given by

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Match the statements in Column I with those in Column II. [Note : Here $z$ takes values in the complex plane and $I m z$ and $R ez$ denote , respectively, the imaginary part and the real part of $z$ . ]

List-I

List-II

If $z$ be a complex number satisfying $∣ Re (z) ∣ + ∣ Im (z) ∣ = 4$ , then $∣ z ∣$ cannot be:

If $z$ be a complex number satisfying $∣ R e (z) ∣ + ∣ I m (z) ∣ = 4$ , then $∣ z ∣$ cannot be:

Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The inequality ∣z−4∣<∣z−2∣ represents the region given by