Let z be a complex number such that the real part of (z-2i)/(z+2i) is zero. Then, the maximum value

Q: Wait, why do we use $w + \bar{w} = 0$ instead of just substituting $z = x + iy$?

You absolutely could substitute $z = x + iy$, but that path is fraught with tedious algebraic expansion. Using the property that a complex number is purely imaginary if and only if $w + \bar{w} = 0$ is a powerful, elegant shortcut that often leads to the answer much faster in JEE-level problems. It keeps the geometry clean and helps you avoid calculation errors.

Q: Why is the maximum distance $OP + r$?

Think of the circle as a set of points at a fixed distance from the origin. If you have a point $P$ outside the circle, the shortest path to the circle is along the line segment connecting $P$ to the center, minus the radius. Conversely, the longest path is along that same line, but extending to the far side of the circle, which is the distance to the center plus the radius. It's all about the geometry of the line passing through the center!

Q: Is it possible for the denominator to be zero?

Great catch! The expression $\frac{z-2i}{z+2i}$ is undefined if $z = -2i$. However, our locus is $|z| = 2$, which means $z$ can be any point on the circle of radius 2. While $-2i$ is on this circle, the maximum distance occurs at the point diametrically opposite to the point on the circle closest to $P$. Since the maximum distance is 12, we are clearly not at the undefined point, so the solution remains valid.

Question

Let $z$ be a complex number such that the real part of $\frac{z-2i}{z+2i}$ is zero. Then, the maximum value of $|z - (6 + 8i)|$ is equal to

Options:

(A) 12

(B) 10

(C) 8

(D) $\infty$

Sigma Percentile · Accepted Answer

### The Dance of Complex Numbers: Unveiling the Geometry

Welcome, future engineers! Today, we are going to peel back the layers of a beautiful complex number problem.

Often, students see complex numbers and immediately reach for $z = x + iy$. While that is a valid tool, the true mastery of JEE Advanced lies in recognizing when to use the geometric and algebraic properties of complex numbers to bypass the grunt work. Let us embark on this journey.

### Phase 1: The Algebraic Trap

We are given the condition $	ext{Re}\left( \frac{z-2i}{z+2i} ight) = 0$. The moment you see a real part of a fraction being zero, your mind should immediately jump to the definition of a purely imaginary number.

If a complex number $w$ has a real part of zero, it means $w$ is purely imaginary. The defining characteristic of a purely imaginary number is that $w + \bar{w} = 0$. This is our golden key.

Let $w = \frac{z-2i}{z+2i}$. Our condition becomes:

$$ \frac{z-2i}{z+2i} + \overline{\left(\frac{z-2i}{z+2i}ight)} = 0 $$

When we take the conjugate of the fraction, we apply it to both the numerator and the denominator:

$$ \frac{z-2i}{z+2i} + \frac{\bar{z}+2i}{\bar{z}-2i} = 0 $$

By taking the least common multiple and cross-multiplying, the denominators effectively vanish because they multiply with zero on the right-hand side. We are left with:

$$ (z-2i)(\bar{z}-2i) + (\bar{z}+2i)(z+2i) = 0 $$

### Phase 2: The Geometric Revelation

Now, let us expand this carefully. The first term gives us $z\bar{z} - 2iz - 2i\bar{z} - 4$, and the second term gives us $z\bar{z} + 2iz + 2i\bar{z} - 4$.

When we add them, the terms involving $i$—specifically $-2iz - 2i\bar{z}$ and $+2iz + 2i\bar{z}$—cancel out perfectly! We are left with:

$$ 2z\bar{z} - 8 = 0 \Rightarrow z\bar{z} = 4 $$

Since $z\bar{z} = |z|^2$, we have $|z|^2 = 4$, or simply $|z| = 2$.

This is the moment of revelation. The equation $|z| = 2$ is not just an algebraic result; it is a geometric command. It tells us that $z$ lies on a circle centered at the origin $(0,0)$ with a radius $r = 2$.

### Phase 3: The Distance Challenge

Now, the problem asks for the maximum value of $|z - (6 + 8i)|$. In the Argand plane, $|z - z_0|$ is the distance between the point $z$ and the fixed point $z_0$.

Here, our fixed point is $P(6, 8)$. We are looking for the maximum distance from any point on our circle $|z| = 2$ to the point $P(6, 8)$.

To find this, we connect the center of the circle, which is the origin $O(0,0)$, to the point $P$. The distance $OP$ is calculated using the distance formula:

$$ OP = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 $$

### Phase 4: The Final Leap

Imagine standing at the origin and looking at point $P$, which is 10 units away. The circle is right there, centered at your feet with a radius of 2.

To get to the farthest point on the circle from $P$, you would walk from $P$ through the center $O$ to the other side of the circle. The distance to the center is 10, and you must travel an additional radius of 2 to reach the far edge.

Thus, the maximum distance is:

$$ OP + r = 10 + 2 = \mathbf{12} $$

And there you have it! By combining the algebraic property of conjugates with the geometric interpretation of the modulus, we have turned a potentially daunting problem into a clear, logical path. Keep practicing this synthesis of algebra and geometry—it is the hallmark of a true JEE champion!

Animated Solution for Mathematics - Complex Numbers: Let $z$ be a complex number such that the real part of $\frac{z - 2 i}{z + 2 i}$ is zero. Then, the maximum value of $∣ z - (6 + 8 i) ∣$ is equal to

Select Answer:

Visualized Solution

Understanding the Condition $Re (w) = 0$

Applying the Purely Imaginary Condition

Cross Multiplication

Expanding the Expression

Simplifying to Find the Locus

Geometrical Interpretation of the Locus

Analyzing the Target Expression

Plotting Point $P$

Distance from Origin to $P$

Calculating $O P$

The Maximum Distance Formula

Final Calculation

Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

The Dance of Complex Numbers

Phase 1

Phase 2

Phase 3

Phase 4

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