write the standard form (-1, 3), (4, -3)

Question

Accepted Answer

# Explanation:
## Step1: Identify the operation
We need to find the standard form of the complex number multiplication $(-1 + 3i)(4 - 3i)$. We use the distributive property (FOIL method) for multiplying binomials.
$$
\begin{align*}
(-1 + 3i)(4 - 3i)&=-1\times4+(-1)\times(-3i)+3i\times4+3i\times(-3i)\
\end{align*}
$$
## Step2: Multiply each term
Calculate each product:
$$
\begin{align*}
-1\times4&=-4\
(-1)\times(-3i)&=3i\
3i\times4&=12i\
3i\times(-3i)&=-9i^{2}
\end{align*}
$$
## Step3: Combine like terms and use $i^{2}=-1$
Combine the real and imaginary parts. Remember that $i^{2} = - 1$, so $-9i^{2}=-9\times(-1) = 9$.
$$
\begin{align*}
&-4 + 3i+12i + 9\
=&(-4 + 9)+(3i + 12i)\
=&5+15i
\end{align*}
$$

# Answer:
The standard form is $5 + 15i$

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QUESTION IMAGE

Question

Explanation:

Step1: Identify the operation

Step2: Multiply each term

Step3: Combine like terms and use \(i^{2}=-1\)

Answer: