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Complex Numbers question

2021 · 1 Sep · Shift 2 · Q40
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Complex Numbers question

2021 · 1 Sep · Shift 2 · Q40

JEE MainMathematicsComplex NumbersNumerical+4 / −1
If for the complex numbers z satisfying | z −-− 2 −-− 2i |≤\le≤ 1, the maximum value of | 3iz + 6 | is attained at a + ib, then a + b is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 5

  1. We need to maximize ∣3iz+6∣|3iz+6|∣3iz+6∣ subject to ∣z−2−2i∣≤1.|z-2-2i|\le 1.∣z−2−2i∣≤1.

  2. First simplify the expression: ∣3iz+6∣=3∣iz+2∣.|3iz+6|=3|iz+2|.∣3iz+6∣=3∣iz+2∣. Since multiplication by iii preserves modulus, ∣iz+2∣=∣i(z−2i)∣?|iz+2|=|i(z-2i)|?∣iz+2∣=∣i(z−2i)∣? A cleaner way is to write iz+2=i(z−2i),iz+2=i(z-2i),iz+2=i(z−2i), because i(z−2i)=iz−2i2=iz+2.i(z-2i)=iz-2i^2=iz+2.i(z−2i)=iz−2i2=iz+2. Hence ∣iz+2∣=∣i∣∣z−2i∣=∣z−2i∣.|iz+2|=|i||z-2i|=|z-2i|.∣iz+2∣=∣i∣∣z−2i∣=∣z−2i∣. Therefore ∣3iz+6∣=3∣z−2i∣.|3iz+6|=3|z-2i|.∣3iz+6∣=3∣z−2i∣.

So maximizing ∣3iz+6∣|3iz+6|∣3iz+6∣ is equivalent to maximizing ∣z−2i∣.|z-2i|.∣z−2i∣.

  1. The condition ∣z−(2+2i)∣≤1|z-(2+2i)|\le 1∣z−(2+2i)∣≤1 represents the closed disk centered at C=2+2iC=2+2iC=2+2i with radius 111.

We want the point in this disk farthest from P=2i.P=2i.P=2i.

  1. Compute the distance between the center CCC and the point PPP: CP=∣(2+2i)−2i∣=∣2∣=2.CP=|(2+2i)-2i|=|2|=2.CP=∣(2+2i)−2i∣=∣2∣=2.

For a disk of radius 111, the farthest point from PPP lies on the boundary in the direction away from PPP through the center.

So the maximum distance is CP+1=2+1=3,CP+1=2+1=3,CP+1=2+1=3, and the point where it occurs is z=C+C−P∣C−P∣⋅1.z=C+\frac{C-P}{|C-P|}\cdot 1.z=C+∣C−P∣C−P​⋅1.

Now, C−P=(2+2i)−2i=2,C-P=(2+2i)-2i=2,C−P=(2+2i)−2i=2, so C−P∣C−P∣=1.\frac{C-P}{|C-P|}=1.∣C−P∣C−P​=1. Hence z=(2+2i)+1=3+2i.z=(2+2i)+1=3+2i.z=(2+2i)+1=3+2i.

Thus, a=3,b=2.a=3,\quad b=2.a=3,b=2. Therefore, a+b=5.a+b=5.a+b=5.

  1. Verification: At z=3+2iz=3+2iz=3+2i, ∣z−(2+2i)∣=∣1∣=1,|z-(2+2i)|=|1|=1,∣z−(2+2i)∣=∣1∣=1, so it satisfies the constraint. Also this is indeed the farthest point of the disk from 2i2i2i, hence gives the maximum.

Therefore the required integer is 5.\boxed{5}.5​.

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