Minimize the complex function

Taking $z = x + iy$ , and replacing in the above expression, one gets:

$f(x,y) = \sqrt{(x^3 -3xy^2 + 1)^2 + (3x^2y - y^3)^2} + \sqrt{x^2 + y^2}$

By inspection of the above expression, for most $(x,y) \ne (0,0)$ , the value of $f(x,y)$ is always greater then unity. The function value cannot go below unity, by inspection. Therefore, the minimum value of the function must be unity. To find other pairs for which $f(x,y)$ is unity, I considered points on the unit circle centered around the origin. For such points: $z = \cos{t}+ i\sin{t}$ , such that $0 \le t \le 2\pi$ . Plugging this into $f(z)$ gives f(t), which is, after simplification:

$f(t) = \lvert 2 \cos(1.5 \ t)\rvert + 1$

Now, the above expression is equal to unity when:

$1.5t = \frac{\pi}{2} \implies t = \frac{\pi}{3}$ $1.5t = \frac{3\pi}{2} \implies t = \pi$ $1.5t = \frac{5\pi}{2} \implies t = \frac{5\pi}{3}$

This gives us three more points besides the origin, making a total of $4$ points. For any general points other than the origin and those on the unit circle: $z = x + iy$ , one needs to compute partial derivatives and equate them to zero and solve the resulting system of nonlinear equations, and finally compute the Hessian matrix to check whether the solutions correspond to a maxima and minima. This exercise is tedious. However, for any general points, I can see by inspection that solutions besides origin and the three points on the unit circle are unlikely. Proving this rigorously is what I am unable to do. Looking forward to seeing more solutions.

Building upon Karan's rigorous (and splendiferous) solution, the smallest sum of two moduli is obviously zero. However, $|z|$ and $|z^3+1|$ can never be simultaneously zero. The best one can achieve is to have one of these moduli equal to zero in order to attain the minimum of $f(z).$

Let $z=a+bi$ . Clearly $|z| = 0 \Rightarrow z = 0 + 0i = 0$ . For $|z^3+1| = 0,$ we have:

$-1 = (a+bi)^3 = a^3 + 3a^2bi -3ab^2 - b^3i$ ;

or $a^3 - 3ab^2 = -1$ AND $3a^2b - b^3=0.$

Taking the imaginary term we obtain $b(3a^2-b^2) = 0 \Rightarrow b = 0, \pm \sqrt{3}a$ . Substituting these values into the real term gives:

$a^3 - 3a(0^2) = -1 \Rightarrow a = -1$ ;

$a^3 -3a(3a^2) = -1 \Rightarrow a = \frac{1}{2}$

We have a total of $\boxed{4}$ critical points: $\boxed{z = 0, -1, \frac{1}{2} \pm \frac{\sqrt{3}}{2}i}$ , each of which yields $f_{MIN} = 1$ .