Complexity

Times through by $z^{4}-1$ to get $z^{4}+1 = \frac{i}{\sqrt{3}} (z^{4}-1)$ . Rearranging gives $z^{4}(1- \frac{i}{\sqrt{3}}) = 1- \frac{i}{\sqrt{3}}$ .

From here you can divide through by $1- \frac{i}{\sqrt{3}}$ giving $z^{4}=- \frac{1}{2}- \frac{\sqrt{3}}{2}$ .

If we write z in the form $cos( \frac{2}{3}\pi)+isin(\frac{2}{3}\pi)$ we can easily find the principle root by dividing the argument by 4 (the root we're taking) Finding the other roots is as simple as adding pi/2 to the argument each time. This gives four values of z: $\frac{\sqrt{3}}{2} + \frac{1}{2}i$ , $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and their negatives.

The values of a,b,c and d are therefore $\frac{\sqrt{3}}{2}, \frac{1}{2}, -\frac{1}{2} and \frac{\sqrt{3}}{2}$ respectively. The bracket simplifies to $\sqrt{3}$ which squared is 3