Let's start by calculating the values of $z_1$, $z_2$, and $z_3$, and then proceed to find the required expressions in both polar and standard forms.

Step 1: Calculate $z_1$, $z_2$, and $z_3$

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

Simplify $z_1$ by multiplying the numerator and denominator by the conjugate of the denominator:

$z_1 = \frac{-i}{-1 + i} \cdot \frac{-1 - i}{-1 - i} = \frac{i(-1 - i)}{(-1)^2 - (i)^2} = \frac{-i - i^2}{1 + 1} = \frac{-i + 1}{2} = \frac{1 - i}{2}$

For $z_2 = \frac{1 + i}{1 - i}$:

Simplify $z_2$ similarly by multiplying the numerator and denominator by the conjugate of the denominator:

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

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

Simplify step by step:

1. Calculate $(1 - i)\overline{(1 - i)}$:

$(1 - i)\overline{(1 - i)} = (1 - i)(1 + i) = 1^2 - i^2 = 1 + 1 = 2$

2. Calculate $2(i - 1)i$:

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

3. Calculate $(-i + \sqrt{3})^3$:

First expand $(-i + \sqrt{3})^3$:

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