# How can I find minimal modulus of $z^2 - w z - 4$?

Let $$z$$ and $$w$$ be two complex numbers satisfying conditions: $$|z| = 2$$ and $$|w i - 2 + 5i| = 1.$$ How can I find the minimum of $$|z^2 - w z - 4|$$?

I tried with setting $$z = a + b i$$ and $$w = x + y i$$:

Minimize[
ComplexExpand /@ {Abs[(a + b  I)^2 - (x + I  y) (a + b  I) - 4 ],
Abs[a + b  I] == 2 && Abs[I  (x + I  y) - 2 + 5 I] == 1}, {x, y,
a, b}]

However, I can not get the result.

## 2 Answers

• $$|z|=2$$ means $$z=2e^{i\alpha}$$
• $$|i w-2+5i|=1$$ means $$i w-2+5i=e^{i\beta}$$

From these,

z = 2 E^(I*ɑ);
w = (E^(I*β) + 2 - 5 I)/I;

Abs[z^2 - w z - 4]^2 // ComplexExpand // FullSimplify
(*    -8 (-19 + 4 Cos[2ɑ] - 2 Cos[β] - 4 (2 + Cos[β]) Sin[ɑ] + 5 Sin[β])    *)

Minimize[%, {ɑ, β}, Reals]
(*    {64, {ɑ -> -293π/6, β -> -19π/2}}    *)

Concretely,

{z, w} /. {ɑ -> -293π/6, β -> -19π/2} // ComplexExpand
(*    {-I - Sqrt[3], -4 - 2 I}    *)

Geometric validation: there are two minima $$z=-i\pm\sqrt{3}$$ and $$w=-4-2i$$,

DensityPlot[Sqrt[-8 (-19 + 4 Cos[2ɑ] - 2 Cos[β] - 4 (2 + Cos[β]) Sin[ɑ] + 5 Sin[β])],
{ɑ, 0, 2π}, {β, 0, 2π},
MeshFunctions -> {#3 &}, Mesh -> {{8.01}}, MeshStyle -> White,
PlotPoints -> 100]

• Reals in Minimize[%, {ɑ, β}, Reals] is superfluous. Commented Apr 26 at 7:33

Let us put z=2*Exp[I*t] and I*w - 2 +5*I = Exp[I*s]. The the task can be written as

Minimize[{ComplexExpand[Abs[(2*Exp[I*t])^2 - (Exp[I*s] + 2 - 5*I)/I *2*
Exp[I*t] - 4]^2], t > -Pi && t <= Pi && s > -Pi && s <= Pi}, {s, t}]

{64, {s -> \[Pi]/2, t -> -((5 \[Pi])/6)}}