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

We have the set of complex numbers $$z$$ that satisfy $$|z-2-2i|\le 1$$. By definition of the modulus, this represents all the points whose distance from the fixed point $$2+2i$$ is at most $$1$$. In geometric language, it is the closed circle (disc) with centre $$C=2+2i$$ and radius $$r=1$$.

The expression whose maximum modulus we have to study is $$|3iz+6|$$. To simplify the notation, let us introduce a new complex variable $$w$$ given by

$$w=3iz+6.$$

This is a linear (affine) transformation of the complex plane. Such a map has two parts:

1. Multiplication by $$3i$$, which multiplies every length by $$|3i|=3$$ and rotates the plane by $$90^\circ$$, because multiplication by $$i$$ corresponds to a rotation of $$\frac{\pi}{2}$$ radians counter-clockwise.
2. Translation by $$6$$, which shifts every point $$6$$ units along the real axis.

Because linear maps take circles to circles (or possibly to lines when the radius is zero), the image of our original circle will again be a circle. We now compute its centre and radius one algebraic step at a time.

Let us write $$z$$ in the form $$z=(2+2i)+(z-(2+2i))$$. Substituting this into $$w$$ we obtain

$$\begin{aligned} w &=3i\Bigl[(2+2i)+(z-(2+2i))\Bigr]+6 \\ &=3i(2+2i)+3i\bigl(z-(2+2i)\bigr)+6. \end{aligned}$$

First we evaluate the constant term $$3i(2+2i)+6$$:

$$\begin{aligned} 3i(2+2i) &=3i\cdot2+3i\cdot2i\\ &=6i+6i^{2}\\ &=6i+6(-1)\\ &=-6+6i. \end{aligned}$$

Adding the real shift $$+6$$ gives

$$(-6+6i)+6 = 0+6i = 6i.$$

Hence the image circle has centre

$$O_w = 6i.$$

Next we look at the variable part $$3i\bigl(z-(2+2i)\bigr)$$. The factor $$3i$$ multiplies every distance by $$3$$, so the radius of the image circle is

$$R = 3\cdot r = 3\cdot1 = 3.$$

Therefore the set of all possible $$w$$ is the closed disc

$$|w-6i|\le 3,$$

i.e. a circle centred on the positive imaginary axis at height $$6$$, with radius $$3$$.

We are asked for the maximum possible value of $$|w|=|3iz+6|$$. Geometrically, $$|w|$$ is simply the distance of the point $$w$$ from the origin. Inside a fixed circle, the farthest point from the origin always lies on the boundary along the line that joins the origin to the circle’s centre. So we move from the centre $$6i$$ further straight outward by one radius.

The distance of the centre from the origin is

$$|6i|=6.$$

Adding one radius gives the maximal possible modulus:

$$|w|_{\max}=6+3=9.$$

The exact point on the circle where this maximum is reached is obtained by moving from $$6i$$ in the same (purely imaginary) direction by length $$3$$:

$$w_{\max}=6i+3i=9i.$$

Thus the extremal value occurs at the complex number

$$w_{\max}=9i.$$

To find the corresponding $$z$$ we solve the linear equation $$w=3iz+6$$ with $$w=9i$$:

$$\begin{aligned} 3iz+6 &=9i\\ 3iz &=9i-6\\ &=-6+9i. \end{aligned}$$

Now we divide by $$3i$$. The reciprocal of $$i$$ is $$-i$$ because $$i\cdot(-i)=1$$. Proceeding step by step,

$$\begin{aligned} z &=\frac{-6+9i}{3i}\\[4pt] &=\frac{3(-2+3i)}{3i}\\[4pt] &=-\frac{2-3i}{i}\\[4pt] &=(-2+3i)\cdot(-i)\quad(\text{since }1/i=-i)\\[4pt] &=2i+3(-i^2)\\[4pt] &=2i+3(1)\\[4pt] &=3+2i. \end{aligned}$$

Thus the extremal point in the original $$z$$-plane is

$$z_{\max}=3+2i.$$

Comparing with the statement of the problem, this is $$a+ib$$ with $$a=3$$ and $$b=2$$. Therefore

$$a+b = 3+2 = 5.$$

So, the answer is $$5$$.