1.

If \( C \cdot N \) satisfying \( |z-2-2 i| \leqslant 1 \), the maxe value of \( |3 i z+6| \) is attainted at \( a+i b \), then \( a+b= \)

Answer»

Let \(z = x + iy\)

Then

\(|z - 2 - 2i|\le 1\)

⇒ \(|(x - 2) + i (y - 2)| \le 1\)

⇒ \(|(x - 2) + i(y - 2)|^2 \le 1\)

⇒ \((x - 2)^2 + (y - 2)^2 \le 1\)    .....(1)

which represent the interior of a circle whose radius is 1 unit and centre is (2, 2).

Now,

\(|3iz + 6|^2 = |3i (x + iy) + 6|^2\)

\(=|3xi - 3y + 6|\)

\(= 9x^2 + (6 - 3y) ^2\)

\(= 9x^2 + 9(y - 2)^2\)

\(= 9x^2 + 9(1 - (x -2)^2)\)    (From (1))

\(= 9x^2 + 9 - 9 (x^2 - 4x + 4)\)

\(= -27 + 36 x \) but \(-1\le x -2\le 1 \) or \(1 \le x \le 3\)

\(|3iz + 6|\) is maximum if x = 3

Then \((y - 2)^2 = 1 - (3 -2)^2 = 1 - 1=0\) 

⇒ \(y = 2\)

Hence, maximum value of \(|3iz + 6|\) is 81 attained at \(3 +2i\).

\(\therefore a + b = 3 + 2 = 5\).


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