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MathMedAdv 2020 • Paper 2

Minimum Distance Squared Between Complex Numbers in Set S

For a complex number zz, let Re(z)\text{Re}(z) denote its real part. Let SS be the set containing all complex numbers zz satisfying z4−∣z∣4=4iz2z^4 - |z|^4 = 4 i z^2, where i=−1i = \sqrt{-1}. Find the minimum possible value of ∣z1−z2∣2|z_1 - z_2|^2, where z1,z2∈Sz_1, z_2 \in S with Re(z1)>0\text{Re}(z_1) > 0 and Re(z2)<0\text{Re}(z_2) < 0.

Official Numerical Answer8

Topics & Concepts

Complex Numbers and Quadratic EquationsComplex Numbers

Step-by-Step Solution

To find the minimum possible value of ∣z1−z2∣2|z_1 - z_2|^2 for z1,z2∈Sz_1, z_2 \in S with Re(z1)>0\text{Re}(z_1) > 0 and Re(z2)<0\text{Re}(z_2) < 0, we first determine the set SS defined by the equation: z4−∣z∣4=4iz2z^4 - |z|^4 = 4 i z^2

Let z=x+iyz = x + iy, where x,y∈Rx, y \in \mathbb{R}. Then: z2=(x2−y2)+2ixyz^2 = (x^2 - y^2) + 2ixy z4=(x2−y2)2−4x2y2+4ixy(x2−y2)z^4 = (x^2 - y^2)^2 - 4x^2 y^2 + 4ixy(x^2 - y^2) ∣z∣4=(x2+y2)2=(x2−y2)2+4x2y2|z|^4 = (x^2 + y^2)^2 = (x^2 - y^2)^2 + 4x^2 y^2

Substituting these expressions into the given equation: LHS=z4−∣z∣4=−8x2y2+4ixy(x2−y2)\text{LHS} = z^4 - |z|^4 = -8x^2 y^2 + 4ixy(x^2 - y^2) RHS=4iz2=4i((x2−y2)+2ixy)=−8xy+4i(x2−y2)\text{RHS} = 4i z^2 = 4i((x^2 - y^2) + 2ixy) = -8xy + 4i(x^2 - y^2)

Equating the real and imaginary parts of both sides gives two simultaneous equations:

  1. Real parts: −8x2y2=−8xy  ⟹  8xy(xy−1)=0-8x^2 y^2 = -8xy \implies 8xy(xy - 1) = 0
  2. Imaginary parts: 4xy(x2−y2)=4(x2−y2)  ⟹  4(xy−1)(x2−y2)=04xy(x^2 - y^2) = 4(x^2 - y^2) \implies 4(xy - 1)(x^2 - y^2) = 0

For z1z_1 and z2z_2, we are given that Re(z1)>0\text{Re}(z_1) > 0 and Re(z2)<0\text{Re}(z_2) < 0, which implies x≠0x \neq 0. If xy=0xy = 0, then y=0y = 0, which from the imaginary part equation leads to x2=0  ⟹  x=0x^2 = 0 \implies x = 0, yielding z=0z = 0, which contradicts Re(z)≠0\text{Re}(z) \neq 0.

Therefore, we must have: xy−1=0  ⟹  xy=1xy - 1 = 0 \implies xy = 1

Thus, non-zero complex numbers z∈Sz \in S lie on the hyperbola xy=1xy = 1.

Let z1=x1+iy1z_1 = x_1 + i y_1 and z2=x2+iy2z_2 = x_2 + i y_2. Since xy=1xy = 1:

  • For z1z_1: Re(z1)=x1>0  ⟹  y1=1x1\text{Re}(z_1) = x_1 > 0 \implies y_1 = \frac{1}{x_1}
  • For z2z_2: Re(z2)=x2<0  ⟹  y2=1x2\text{Re}(z_2) = x_2 < 0 \implies y_2 = \frac{1}{x_2}

Let x1=a>0x_1 = a > 0 and x2=−bx_2 = -b with b>0b > 0. Then: z1=a+ia,z2=−b−ibz_1 = a + \frac{i}{a}, \quad z_2 = -b - \frac{i}{b}

The squared distance between z1z_1 and z2z_2 is: ∣z1−z2∣2=(a+b)2+(1a+1b)2|z_1 - z_2|^2 = (a + b)^2 + \left(\frac{1}{a} + \frac{1}{b}\right)^2 ∣z1−z2∣2=(a+b)2+(a+b)2a2b2=(a+b)2(1+1a2b2)|z_1 - z_2|^2 = (a + b)^2 + \frac{(a + b)^2}{a^2 b^2} = (a + b)^2 \left(1 + \frac{1}{a^2 b^2}\right)

By the AM-GM inequality, (a+b)≥2ab(a + b) \ge 2\sqrt{ab}, which gives (a+b)2≥4ab(a + b)^2 \ge 4ab. Setting u=ab>0u = ab > 0: ∣z1−z2∣2≥4ab(1+1a2b2)=4u(1+1u2)=4(u+1u)|z_1 - z_2|^2 \ge 4ab \left(1 + \frac{1}{a^2 b^2}\right) = 4u \left(1 + \frac{1}{u^2}\right) = 4 \left(u + \frac{1}{u}\right)

By applying the AM-GM inequality again to u+1uu + \frac{1}{u}: u+1u≥2u + \frac{1}{u} \ge 2

Therefore: ∣z1−z2∣2≥4×2=8|z_1 - z_2|^2 \ge 4 \times 2 = 8

Equality holds when a=ba = b and u=ab=1u = ab = 1, which gives a=1a = 1 and b=1b = 1. In this case, z1=1+iz_1 = 1 + i and z2=−1−iz_2 = -1 - i, giving: ∣z1−z2∣2=∣2+2i∣2=22+22=8|z_1 - z_2|^2 = |2 + 2i|^2 = 2^2 + 2^2 = 8

Hence, the minimum possible value of ∣z1−z2∣2|z_1 - z_2|^2 is 8.

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