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(1−4i)(−1−1i)

Let's solve the expression \((1 - 4i)(-1 - 1i)\), where \(i\) represents the imaginary unit.

We can apply the distributive property (FOIL method) to expand the expression:

\[
(1 - 4i)(-1 - 1i) = 1 \cdot (-1) + 1 \cdot (-1i) + (-4i) \cdot (-1) + (-4i) \cdot (-1i)
\]

Now, let's calculate each term:

1. \(1 \cdot (-1) = -1\)
2. \(1 \cdot (-1i) = -i\)
3. \((-4i) \cdot (-1) = 4i\)
4. \((-4i) \cdot (-1i) = 4i^2 = 4(-1) = -4\) (since \(i^2 = -1\))

Now, let's combine all the terms:

\[
-1 - i + 4i - 4
\]

Group the real and imaginary parts:

\[
(-1 - 4) + (-i + 4i) = -5 + 3i
\]

So, the result is:

\[
\boxed{-5 + 3i}
\]

Would you like more details on how this was done or have any other questions?

Here are 5 related questions to practice:

1. Simplify \((3 + 2i)(2 - i)\).
2. Expand \((5 - i)(-3 + 4i)\).
3. What is the conjugate of \(2 + 3i\)?
4. Find the magnitude of \(4 + 7i\).
5. Simplify \((1 + i)^2\).

**Tip:** When multiplying complex numbers, remember to treat \(i^2\) as \(-1\).

Learn how to multiply complex numbers with the problem (1 - 4i)(-1 - 1i). This step-by-step solution explores the distributive property, handling imaginary units, and simplifying the result to -5 + 3i. Ideal for students in Grades 10-12.

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