If for the complex numbers z satisfying | z - 2 - 2i | ≤ 1 , the...

Given that: $| z - 2 - 2i | \le 1$ which represents a circle with centre at $( 2, 2 ) and radius$= 1$.</p><p>Now$| 3iz + 6 | = | 3i ( z - 2i ) | = | 3i | | z - 2i |$</p><p>As we know that, \left[| {z}_{1}{z}_{2} | = | {z}_{1} | | {z}_{2} |\right]$

From tha diagram we can see that, $| 3iz + 6 | = 3 | z - 2i |$ is maximum at $z = 3 + 2i$ because a point on the circle, which has maximum distance from $( 0, 2 )$ is $( 3, 2 )$.

So, $a = 3 \text{\text{ and } }$ b = 2

$⇒ a + b = 5$