To solve the problem, we need to first express $z_1$, $z_2$, and $z_3$ in their standard forms (rectangular forms) and then compute the required expressions.

Step 1: Simplify $z_1 = \frac{-i}{-1 + i}$

Simplify the expression by multiplying the numerator and the denominator by the conjugate of the denominator:

$z_1 = \frac{-i}{-1 + i} \times \frac{-1 - i}{-1 - i} = \frac{-i(-1 - i)}{(-1 + i)(-1 - i)}$

The denominator is: $(-1 + i)(-1 - i) = 1 + 1 = 2$

The numerator is: $-i(-1 - i) = i + 1$

So, $z_1 = \frac{1 + i}{2} = \frac{1}{2} + \frac{i}{2}$

Step 2: Simplify $z_2 = \frac{1+i}{1-i}$

Similarly, simplify by multiplying by the conjugate of the denominator:

$z_2 = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{(1 + i)^2}{(1 - i)(1 + i)} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i$

Step 3: Simplify $z_3 = \frac{1}{10} \left[ 2(i - 1)i + (-i + \sqrt{3})^3 \right] + (1-i)\overline{(1-i)}$

First, simplify each term:

$2(i-1)i = 2i(i-1) = 2i^2 - 2i = -2 - 2i$

For the cubic term, $(-i + \sqrt{3})^3$:

Let $w = -i + \sqrt{3}$. Then: $w^2 = (-i + \sqrt{3})^2 = (-i)^2 + 2(-i)(\sqrt{3}) + (\sqrt{3})^2 = -1 - 2i\sqrt{3} + 3 = 2 - 2i\sqrt{3}$

$w^3 = w^2(-i + \sqrt{3}) = (2 - 2i\sqrt{3})(-i + \sqrt{3})$

Expanding:

= -2i + 2\sqrt{3} - 6 + 2i^2 = -2i + 2\sqrt{3} - 6 + 2i(-i) + 6\sqrt{3} = (-6 + 2i + 6) + (2i - 6i) = -2\sqrt{3}i$$ So: $$(-i + \sqrt{3})^3 = -6 + 2\sqrt{3} + i(-2 - 6i)$$ Adding the above with $$-2 - 2i + (-6 + 2\sqrt{3}) + i(\overline{1 - i})$$ we add the following: Conjugate: $$\overline{(1 - i)} = 1 + i$$: $$= (-6 + 2\sqrt{3}) =$$ equals: ### Expression forms: Express in simplified: $$z_3 = -6 + 2\sqrt{3} + (1+i)(1-i)$$ Combining all of these to a real simplified expression on both $$/bar(z_1z_2)/z_3,z_1barz_2/z3$$. Tip: Solve math on both formats by steps.