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JEE Main 2024
Complex Numbers
Complex Numbers
Easy

Question

Let the minimum value v0v_{0} of v=z2+z32+z6i2,zCv=|z|^{2}+|z-3|^{2}+|z-6 i|^{2}, z \in \mathbb{C} is attained at z=z0{ }{z}=z_{0}. Then 2z02zˉ03+32+v02\left|2 z_{0}^{2}-\bar{z}_{0}^{3}+3\right|^{2}+v_{0}^{2} is equal to :

Options

Solution

Key Concepts and Formulas

  • Modulus of a Complex Number: If z=x+iyz = x + iy, where xx and yy are real numbers, then z=x2+y2|z| = \sqrt{x^2 + y^2}, and z2=x2+y2|z|^2 = x^2 + y^2.
  • Completing the Square: The method of completing the square transforms a quadratic expression of the form ax2+bx+cax^2 + bx + c into a(xh)2+ka(x - h)^2 + k, where h=b2ah = -\frac{b}{2a} and k=cb24ak = c - \frac{b^2}{4a}.
  • Centroid of Complex Numbers: The centroid of complex numbers z1,z2,,znz_1, z_2, \ldots, z_n is given by z=z1+z2++znnz = \frac{z_1 + z_2 + \cdots + z_n}{n}. The sum k=1nzzk2\sum_{k=1}^n |z - z_k|^2 is minimized when zz is the centroid.

Step-by-Step Solution

Step 1: Convert to Cartesian Coordinates

We are given v=z2+z32+z6i2v = |z|^2 + |z-3|^2 + |z-6i|^2. To work with this algebraically, we express zz in Cartesian form: Let z=x+iyz = x + iy, where xx and yy are real numbers. This allows us to express the modulus in terms of xx and yy.

Now we expand each term:

  • z2=x+iy2=x2+y2|z|^2 = |x + iy|^2 = x^2 + y^2
  • z32=(x3)+iy2=(x3)2+y2=x26x+9+y2|z - 3|^2 = |(x - 3) + iy|^2 = (x - 3)^2 + y^2 = x^2 - 6x + 9 + y^2
  • z6i2=x+i(y6)2=x2+(y6)2=x2+y212y+36|z - 6i|^2 = |x + i(y - 6)|^2 = x^2 + (y - 6)^2 = x^2 + y^2 - 12y + 36

Step 2: Expand and Simplify the Expression for vv

Substitute these expanded forms back into the expression for vv: v=(x2+y2)+(x26x+9+y2)+(x2+y212y+36)v = (x^2 + y^2) + (x^2 - 6x + 9 + y^2) + (x^2 + y^2 - 12y + 36) Combine like terms: v=3x2+3y26x12y+45v = 3x^2 + 3y^2 - 6x - 12y + 45

Step 3: Complete the Square to Find the Minimum Value

We complete the square for the xx and yy terms to find the minimum value. First, factor out the coefficient 33: v=3(x22x)+3(y24y)+45v = 3(x^2 - 2x) + 3(y^2 - 4y) + 45

Complete the square for x22xx^2 - 2x: x22x=(x1)21x^2 - 2x = (x - 1)^2 - 1

Complete the square for y24yy^2 - 4y: y24y=(y2)24y^2 - 4y = (y - 2)^2 - 4

Substitute these back into the expression for vv: v=3((x1)21)+3((y2)24)+45v = 3((x - 1)^2 - 1) + 3((y - 2)^2 - 4) + 45 v=3(x1)23+3(y2)212+45v = 3(x - 1)^2 - 3 + 3(y - 2)^2 - 12 + 45 v=3(x1)2+3(y2)2+30v = 3(x - 1)^2 + 3(y - 2)^2 + 30

Step 4: Determine the Minimum Value (v0v_0) and the Complex Number (z0z_0) where it's attained

Since (x1)20(x - 1)^2 \ge 0 and (y2)20(y - 2)^2 \ge 0, the minimum value of vv occurs when x=1x = 1 and y=2y = 2. Thus, z0=x+iy=1+2iz_0 = x + iy = 1 + 2i. The minimum value is v0=3(0)+3(0)+30=30v_0 = 3(0) + 3(0) + 30 = 30.

Step 5: Calculate the Final Expression 2z02zˉ03+32+v02\left|2 z_{0}^{2}-\bar{z}_{0}^{3}+3\right|^{2}+v_{0}^{2}

We have z0=1+2iz_0 = 1 + 2i and v0=30v_0 = 30.

  1. Calculate z02z_0^2: z02=(1+2i)2=1+4i+4i2=1+4i4=3+4iz_0^2 = (1 + 2i)^2 = 1 + 4i + 4i^2 = 1 + 4i - 4 = -3 + 4i

  2. Calculate zˉ03\bar{z}_0^3: zˉ0=12i\bar{z}_0 = 1 - 2i zˉ03=(12i)3=16i+12i28i3=16i12+8i=11+2i\bar{z}_0^3 = (1 - 2i)^3 = 1 - 6i + 12i^2 - 8i^3 = 1 - 6i - 12 + 8i = -11 + 2i

  3. Calculate 2z02zˉ03+32z_0^2 - \bar{z}_0^3 + 3: 2z02zˉ03+3=2(3+4i)(11+2i)+3=6+8i+112i+3=8+6i2z_0^2 - \bar{z}_0^3 + 3 = 2(-3 + 4i) - (-11 + 2i) + 3 = -6 + 8i + 11 - 2i + 3 = 8 + 6i

  4. Calculate 2z02zˉ03+32|2z_0^2 - \bar{z}_0^3 + 3|^2: 8+6i2=82+62=64+36=100|8 + 6i|^2 = 8^2 + 6^2 = 64 + 36 = 100

  5. Calculate v02v_0^2: v02=302=900v_0^2 = 30^2 = 900

  6. Calculate the final expression: 2z02zˉ03+32+v02=100+900=1000|2z_0^2 - \bar{z}_0^3 + 3|^2 + v_0^2 = 100 + 900 = 1000

Common Mistakes & Tips

  • Be careful with signs when expanding and simplifying complex expressions. A common mistake is to incorrectly calculate powers of ii.
  • Remember that the minimum value of a sum of squared distances is attained at the centroid of the points.
  • Double-check your arithmetic when calculating z02z_0^2 and zˉ03\bar{z}_0^3.

Summary

We converted the complex numbers to Cartesian coordinates, completed the square to find the minimum value v0=30v_0 = 30, and the point where it occurs z0=1+2iz_0 = 1 + 2i. Then we computed 2z02zˉ03+32+v02=1000|2 z_{0}^{2}-\bar{z}_{0}^{3}+3|^{2}+v_{0}^{2} = 1000.

The final answer is \boxed{1000}, which corresponds to option (A).

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