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.