Question

1. $(-2i) - 7 - (-2 + 8i)$
2. $(5 - 7i) - (2 + i)$

simplify and write complex number in standard form.

1. $5(-8i)(1 + i)$
2. $(-8 - 8i)^2$
3. $(-4 + 2i)^2$

challenge problems: complete any two and write in standard form.

1. $-6 - (-2 - 2i) - (5 - 4i)$
2. $-4(-7 + 8i)(-5 + 6i)$
3. $(-5 + 5i) - (4 - 2i) + (-8 - 7i)^2$
4. $(-4 - 6i) - (-7 - i) + (-5 - 4i)(8 + 4i)$

rationalize the denominator to find the quotient.

1. $\frac{3i}{-3 + i}$
2. $\frac{10i}{-6 + i}$
3. $\frac{5}{6 - 2i}$

simplify.

1. $\frac{-7 - 3i}{2 + 6i}$
2. $\frac{4 + i}{8 - 7i}$
3. $\frac{7 - 7i}{7 - 5i}$
4. $\frac{-10 - 2i}{-8 + 6i}$

Explanation:

Let's solve problem 7: \( 5(-8i)(1 + i) \)

Step 1: Multiply the constants and the imaginary unit

First, multiply \( 5 \) and \( -8i \).
\( 5 \times (-8i) = -40i \)

Step 2: Distribute the \( -40i \)

Now, distribute \( -40i \) over \( (1 + i) \).
\( -40i(1 + i) = -40i \times 1 + (-40i) \times i \)
\( = -40i - 40i^2 \)

Step 3: Replace \( i^2 \) with \( -1 \)

Recall that \( i^2 = -1 \). Substitute this into the expression.
\( -40i - 40(-1) = -40i + 40 \)

Step 4: Write in standard form

The standard form of a complex number is \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part.
So, rearranging, we get \( 40 - 40i \)

Answer:

\( 40 - 40i \)