Question

Let the minimum value v₀ of
v = |z|²+|z-3|²+|z-6i|²,z∈C
is attained at z = z₀. Then
\(|2z^2_0-\overline{z}^3_0+3|^2+v^2_0\)
is equal to

• A. 1000
• B. 1024
• C. 1105
• D. 1196

Answer 1

The Correct Option is A

To solve the problem, we need to minimize the expression \(v = |z|^2 + |z-3|^2 + |z-6i|^2\) for \(z \in \mathbb{C}\) and then find the value of \(|2z^2_0 - \overline{z}^3_0 + 3|^2 + v^2_0\) where \(z_0\) is the point where \(v\) is minimized.

Let's break it down:

1. Represent the complex number \(z\) as \(z = x + yi\) where \(x\) and \(y\) are real numbers.
2. Then, compute each squared modulus term:
   • \(|z|^2 = x^2 + y^2\)
   • \(|z - 3|^2 = (x-3)^2 + y^2 = x^2 - 6x + 9 + y^2\)
   • \(|z - 6i|^2 = x^2 + (y-6)^2 = x^2 + y^2 - 12y + 36\)
3. Combine these to form the expression for \(v\): \(v = (x^2 + y^2) + (x^2 - 6x + 9 + y^2) + (x^2 + y^2 - 12y + 36)\)
4. Simplify the total expression:
   • \(v = 3x^2 + 3y^2 - 6x - 12y + 45\)
5. Now, to minimize this quadratic expression, complete the square:
   • For \(x\), complete the square: \(3(x^2 - 2x) = 3((x-1)^2 - 1) = 3(x-1)^2 - 3\)
   • For \(y\), complete the square: \(3(y^2 - 4y) = 3((y-2)^2 - 4) = 3(y-2)^2 - 12\)
6. Thus, we have: \(v = 3(x-1)^2 + 3(y-2)^2 - 3 - 12 + 45 = 3(x-1)^2 + 3(y-2)^2 + 30\)
7. The minimum value of \(v\) is clearly at \(x = 1\) and \(y = 2\), giving \(v_0 = 30\).
8. Next, find \(z_0 = 1 + 2i\) where \(v\) attains its minimum value. Compute \(|2z^2_0 - \overline{z}^3_0 + 3|^2\):
   • \(z_0 = 1 + 2i\), so \(z_0^2 = (1 + 2i)^2 = -3 + 4i\)
   • \(2z_0^2 = -6 + 8i\)
   • Conjugate: \(\overline{z}_0 = 1 - 2i\), then \(\overline{z}_0^3 = (1-2i)^3 = -11 + 2i\)
   • \(- \overline{z}_0^3 = 11 - 2i\)
   • Therefore, compute \(2z^2_0 - \overline{z}^3_0 + 3 = (-6 + 8i) + (11 - 2i) + 3 = 8 + 6i\)
   • Finally, \(|8 + 6i|^2 = 8^2 + 6^2 = 64 + 36 = 100\)
9. Therefore, \(|2z^2_0 - \overline{z}^3_0 + 3|^2 + v^2_0 = 100 + 30^2 = 100 + 900 = 1000\).

The value is 1000. Thus, the correct answer is 1000.