To solve $(-2i)(5i)(-i)$, let's proceed step by step:

Step 1: Multiply the coefficients

First, multiply the real coefficients: $-2 \cdot 5 \cdot -1 = 10$

Step 2: Multiply the imaginary parts

Recall that $i^2 = -1$. Therefore: $i \cdot i = i^2 = -1$ $i^3 = i^2 \cdot i = (-1) \cdot i = -i$

Now, combine the imaginary terms step by step: $(-2i)(5i) = -10i^2 = -10(-1) = 10$ $(10)(-i) = -10i$

Final Answer:

$(-2i)(5i)(-i) = -10i$

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Would you like further explanation or have any questions?

Here are 5 related questions for further exploration:

1. What is the value of $i^4$ and how does it simplify complex numbers?
2. How do you simplify $(3 + 2i)(1 - i)$?
3. What happens when you multiply two complex conjugates, such as $(a + bi)(a - bi)$?
4. How can imaginary roots appear in polynomial equations?
5. How does the polar form of a complex number assist in multiplication?

Tip: Always remember $i^2 = -1$, as it's the foundation for simplifying complex number multiplications!