The complex number $$z = \dfrac{i-1}{\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}}$$ is equal to:

We need to simplify $$z = \dfrac{i-1}{\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}}$$. To do this, we first express the numerator in polar form. Since $$i - 1 = -1 + i$$ has modulus $$|i-1| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$$ and argument $$\theta = \pi - \tan^{-1}(1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$, we can write $$i - 1 = \sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)$$.

The denominator is already in polar form: $$\cos\frac{\pi}{3} + i\sin\frac{\pi}{3} = e^{i\pi/3}$$.

Dividing the numerator by the denominator gives $$ z = \frac{\sqrt{2} \cdot e^{i \cdot 3\pi/4}}{e^{i\pi/3}} = \sqrt{2} \cdot e^{i(3\pi/4 - \pi/3)} = \sqrt{2} \cdot e^{i(9\pi/12 - 4\pi/12)} = \sqrt{2} \cdot e^{i \cdot 5\pi/12},$$ so that $$ z = \sqrt{2}\left(\cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12}\right).$$