Ok. I think this might be calculus. real hard.

(a,b) = (1,sqrt(3)).

Hi Nerdiest.  

I just worked out each bit seperately and then put it altogether.

I have not presented it becasue it is a bit long and I no doubt made careless errors 

But I will walk you through it

$z=a+bi\\ z^3=a^3+3a^2bi+3a(bi)^2+(bi)^3=etc\\ |z|^2=a^2+b^2$

Do the addition

The real part is 0 and the imaginary part is 0

Solve simultaneously.

Hi.

Given:                                                    $z=a+bi$

So:                           $z^3=(a+bi)^3=a^3+3a^2bi-3ab^2-b^3i$

Given:                                   $z^3+\left | z\right |^2+z=0$

Substitute $z,z^3$, and we know: $\left |z \right |^2=\sqrt{a^2+b^2}^2=a^2+b^2$

$a^3+3a^2bi-3ab^2-b^3i+(a^2+b^2)+a+bi=0$

We equate real parts = 0 and imaginary parts = 0             (As the right-hand side is zero, this means both the real and imaginary parts are zero.) (Real parts without the "i" and imaginary parts with the "i").

So: Set $Re=0:$  $a^3-3ab^2+a^2+b^2+a=0$       (1)

      Set $Im=0:$   $3a^2b-b^3+b=0$                          (2)

Given b is not equal to zero so, divide (2) by b:

      $3a^2-b^2+1=0$

Let's make b the subject of the equation: (That is, solve for b.)

$b=\sqrt{3a^2+1}$

Hence, substituting b in (1) to get:

$a^3-3a(3a^2+1)+a^2+3a^2+1+a=0$   (Expand.)

$a^3-9a^3-3a+a^2+3a^2+1+a=0$        (Simplify.)

$-8a^3-2a+4a^2+1=0$                               (Multiply by -1)

$8a^3-4a^2+2a-1=0$

We can factor this by grouping:

$4a^2(2a-1)+(2a-1)=0$

$(2a-1)(4a^2+1)=0$

We know that a is a positive real number.

So we only take:  $a=\frac{1}{2}$

Hence, $b=\sqrt{3(0.5)^2+1}=\sqrt{\frac{7}{4}}=\frac{\sqrt{7}}{2}$

Therefore, $(\frac{1}{2},\frac{{\sqrt{7}}}{2})$ is the desired answer.

I hope this helps :)!

Incidentally, this has nothing to do with calculus.