Simply Complex!

Even easier solution:

Note that the two equations are equivalent to: $a^3-3ab^2=3$ $3a^2b-b^3=3$ Multiply the second equation by $i$ and add it to the first to get $a^3+3a^2bi-3ab^2i-b^3=(a+bi)^3=3+3i$

Thus, the modulus of $a+bi$ is equal to $\sqrt[3]{|3+3i|}=\sqrt[3]{\sqrt{18}}\approx \boxed{1.619}$

The hard solution is:

Sum the two equations to obtain:

$a^3+b^3=3ab(a+b)$

Factor the LHS and simplify:

$(a+b)(a^2-ab+b^2)=3ab(a+b)$

$a^2-ab+b^2=3ab \Longrightarrow a^2+b^2=4ab$

Now, let $a=R\cos \theta$ and $b=R\sin \theta$ :

$(R\cos \theta)^2+(R\sin \theta)^2=4(R\cos \theta)(R \sin \theta)$

Simplify and solve for $\theta$ :

$R^2=4R^2 \sin \theta \cos \theta$

$1=2 \sin (2\theta)$

$\theta=15°$

Finally, find the modulus by substituting $\theta$ in any of the two original equations. Let's choose the first one:

$a^3=3(1+ab^2)$

$(R\cos 15°)^3=3(1+(R\cos 15°)(R\sin 15°)^2)$

$R^3(\cos^3 15°-3\cos 15° \sin^2 15°)=3$

$\dfrac{R^3}{\sqrt{2}}=3 \Longrightarrow R^3=3\sqrt{2}=\sqrt{18}$

$R=\sqrt[6]{18} \approx \boxed{1.619}$

Let s = a+b and p = ab. Adding the given equations,

a^3+b^3 = 3ab^2+3(a^2)b,

(a+b)(a^2-ab+b^2) = 3ab(a+b),

(a+b)(a^2-4ab+b^2) = 0,

s(s^2-6p) = 0,

s = 0 or s^2 = 6p.

Assume s = 0. So b = -a. The first given equation becomes

a^3 = 3(1+a^3),

2a^3 = -3,

a = -(3/2)^(1/3).

Therefore b = -a = (3/2)^(1/3); and so |z| = sqrt(a^2+b^2) ~ 1.6189.

Assume s^2 = 6p. Subtracting the given equations,

a^3-b^3 = 3ab^2-3(a^2)b+6,

(a-b)(a^2+ab+b^2) = 3ab(b-a)+6,

(a-b)(a^2+4ab+b^2) = 6,

(a-b)(s^2+2p) = 6,

(a-b)(8p) = 6,

(a-b)^2 * 16p^2 = 9,

16(s^2-4p)p^2 = 9,

32p^3 = 9,

p = (9/32)^(1/3).

Therefore |z| = sqrt(a^2+b^2) = sqrt(s^2-2p) = 2*sqrt(p) ~ 1.6189.

In either case |z| ~ 1.6189.

The key to this solution is the fact that $|z^3|=|z|^3$ . We know that $z^3=a^3+3a^2bi-3ab^2-b^3i$ and by substituting the given equations, we find that $z^3=3+3i$ .

Combining these two statements, we find that $|z|=\sqrt[3]{|z^3|}=\sqrt[6]{3^2+3^2}$ .

i hate complicated answers !!!! i am never confident when i obtain such a result. first i add up the two equations, ending up in a^2+b^2=4ab, then i use this when substracting the two equations,ending up knowing ab,therefor knowing the modulus...also there are two possible answers since -1.61887 works just as fine.