z = -\sqrt{3} + i, \text{ where } i^2 = -1 - Leaving Cert Mathematics - Question 2 - 2017

Given w is of the form: $w = |w|\left(\cos(30°) + i\sin(30°)\right).$
Since |w| = 3: $w = 3\left(\cos(30°) + i\sin(30°)\right) = 3\left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = \frac{3\sqrt{3}}{2} + i\frac{3}{2}.$

Next, we have: $z = -\sqrt{3} + i.$
Now, we calculate zw: $zw = (-\sqrt{3} + i)\left(\frac{3\sqrt{3}}{2} + i\frac{3}{2}\right).$ Using the distributive property: $zw = -\sqrt{3}\cdot\frac{3\sqrt{3}}{2} + (-\sqrt{3})\cdot i\cdot\frac{3}{2} + i\cdot\frac{3\sqrt{3}}{2} + i^2\cdot\frac{3}{2}$

Since $i^2 = -1$, we simplify: $zw = -\frac{9}{2} + (-\frac{3\sqrt{3}}{2})i + \frac{3\sqrt{3}}{2}i - \frac{3}{2}$

Combining like terms: $zw = -\frac{9}{2} - \frac{3}{2} = -6.$ Thus, in simplest form: $t = -6.$