If for the complex numbers z satisfying |z - 2 - 2i| ≤ 1, the maximum value of |3iz + 6| is attained

Q: Why do we factor out $3i$? Can't we just substitute $z = x + iy$?

You absolutely could, but that leads to a messy algebraic nightmare involving square roots and inequalities. In JEE Advanced, we value elegance. Factoring out $3i$ transforms the problem into a geometric one, which is much faster, more intuitive, and significantly less prone to calculation errors.

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

Think of the triangle inequality. The distance from an external point $P$ to any point $z$ on the disk is bounded. The farthest point must lie on the line passing through $P$ and the center $C$. The distance is the segment from $P$ to $C$ plus the radius $r$ of the disk. Any other point on the disk would form a triangle with $P$ and $C$, and by the triangle inequality, the distance would be less than $PC + r$.

Q: Is it possible for the maximum to be at a point other than the boundary?

For a linear distance function like $|z - z_0|$, the maximum will always occur on the boundary of the disk. The interior points are always closer to the external point than the boundary points, so the maximum must lie on the edge.

Question

If for the complex numbers $z$ satisfying $|z - 2 - 2i| \le 1$, the maximum value of $|3iz + 6|$ is attained at $a + i b$, then $a + b$ is equal to .

Sigma Percentile · Accepted Answer

### Analyzing the Landscape

When you see an expression like $|z - (2 + 2i)| \le 1$, do not just see an inequality. See a landscape. This is the equation of a solid disk in the complex plane.

The center $C$ is at $2 + 2i$, which corresponds to the coordinate $(2, 2)$, and the radius $r$ is $1$. This disk is the exclusive territory where our complex number $z$ is allowed to exist.

### The Art of Simplification

Now, let us turn our attention to the objective function: $|3iz + 6|$. We want to isolate $z$ to understand its relationship with other points.

Let us factor out the coefficient of $z$, which is $3i$:

$$|3i(z + \frac{6}{3i})|$$

Now, let us handle that fraction. We know that $\frac{1}{i} = -i$. Therefore, $\frac{6}{3i} = \frac{2}{i} = 2(-i) = -2i$.

Substituting this back, our objective function transforms into $|3i(z - 2i)|$. By the properties of the modulus, where the modulus of a product is the product of the moduli, we can write this as:

$$|3i| \cdot |z - 2i| = 3|z - 2i|$$

### The Geometric Intuition

This is where the magic happens. The expression $|z - 2i|$ represents the distance between our variable point $z$ and a fixed point $P(0, 2)$.

We are no longer doing complex algebra; we are solving a geometry problem. We have a disk centered at $C(2, 2)$ with radius $1$, and we have a point $P(0, 2)$.

We need to find the point $z$ on or inside this disk that is farthest from $P$. To get as far away as possible, you would walk in a straight line from $P$ through the center $C$ and continue until you hit the far edge of the disk.

### The Final Leap

Let us calculate the distance $PC$. Since $P$ is at $(0, 2)$ and $C$ is at $(2, 2)$, they both lie on the horizontal line $y = 2$.

The distance $PC$ is simply the difference in their x-coordinates: $2 - 0 = 2$. The maximum distance from $P$ to any point on the disk is the distance to the center plus the radius:

$$PC + r = 2 + 1 = 3$$

To find the coordinates of this farthest point $z$, we start at $P(0, 2)$ and move $3$ units in the direction of $C$. Since $C$ is to the right of $P$, we move $3$ units along the positive x-axis.

This lands us at $z = (0 + 3, 2) = (3, 2)$. In complex form, this is $z = 3 + 2i$.

The problem defines this point as $a + ib$. Thus, $a = 3$ and $b = 2$. The final step is to find $a + b$:

$$a + b = 3 + 2 = \mathbf{5}$$

Animated Solution for Mathematics - Complex Numbers: If for the complex numbers $z$ satisfying $∣ z - 2 - 2 i ∣ \leq 1$ , the maximum value of $∣3 i z + 6∣$ is attained at $a + ib$ , then $a + b$ is equal to .

Enter Numerical Value:

Visualized Solution

Visualizing the Constraint

Analyzing the Objective Function

Simplifying the Expression

Applying Modulus Properties

Geometric Interpretation of $∣ z - 2 i ∣$

Distance from P to Center C

Finding the Farthest Point

Locating Point z

Final Calculation

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Landscape

The Art of Simplification

The Geometric Intuition

The Final Leap

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