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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
If ∣z+2−i∣=5|z + 2 - i| = 5, then the maximum value of ∣3z+9−7i∣|3z + 9 - 7i| is
A2020correct
B1515
C55
D1616
Solution
Step 1: Rewrite the target expression so that the given quantity z+2−iz + 2 - i appears. We want 3z+9−7i=3(z+2−i)+c3z + 9 - 7i = 3(z + 2 - i) + c for some constant cc. Expand 3(z+2−i)=3z+6−3i3(z + 2 - i) = 3z + 6 - 3i, so:
c=(3z+9−7i)−(3z+6−3i)=3−4ic = (3z + 9 - 7i) - (3z + 6 - 3i) = 3 - 4i
 Therefore 3z+9−7i=3(z+2−i)+(3−4i)3z + 9 - 7i = 3(z + 2 - i) + (3 - 4i). Step 2: Apply the triangle inequality ∣p+q∣≤∣p∣+∣q∣|p + q| \leq |p| + |q|:
∣3z+9−7i∣=∣3(z+2−i)+(3−4i)∣≤3 ∣z+2−i∣+∣3−4i∣|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|z + 2 - i| = 5, and ∣3−4i∣=32+42=25=5|3 - 4i| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5:
∣3z+9−7i∣≤3⋅5+5=15+5=20|3z + 9 - 7i| \leq 3\cdot 5 + 5 = 15 + 5 = 20
 Step 4: The maximum value 2020 is attained when 3(z+2−i)3(z + 2 - i) and (3−4i)(3 - 4i) point in the same direction, which is possible since zz ranges over the whole circle. Hence the maximum is 2020. Correct answer: (1)
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