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Maximum of |3z+9−7i| Given |z+2−i|=5 | Complex JEE

JEE Maths question with a full step-by-step solution.

Question

|z + 2 - i| = 5

, then the maximum value of

|3z + 9 - 7i|

20

correct

15

5

16

Solution

Step 1: Rewrite the target expression so that the given quantity

z + 2 - i

appears. We want

3z + 9 - 7i = 3(z + 2 - i) + c

for some constant

c

. Expand

3(z + 2 - i) = 3z + 6 - 3i

, so:

c = (3z + 9 - 7i) - (3z + 6 - 3i) = 3 - 4i

Therefore

3z + 9 - 7i = 3(z + 2 - i) + (3 - 4i)

. Step 2: Apply the triangle inequality

|p + q| \leq |p| + |q|

|3z + 9 - 7i| = |3(z + 2 - i) + (3 - 4i)| \leq 3\,|z + 2 - i| + |3 - 4i|

Step 3: Substitute the known values. The given is

|z + 2 - i| = 5

, and

|3 - 4i| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5

|3z + 9 - 7i| \leq 3\cdot 5 + 5 = 15 + 5 = 20

Step 4: The maximum value

20

is attained when

3(z + 2 - i)

and

(3 - 4i)

point in the same direction, which is possible since

z

ranges over the whole circle. Hence the maximum is

20

. Correct answer: (1)

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