Question

4(3 - 2i) + (2 + 3i)^2

Explanation:

Step1: Expand the terms

First, expand \(4(3 - 2i)\) and \((2 + 3i)^2\) using the distributive property (for the first term) and the formula \((a + b)^2 = a^2 + 2ab + b^2\) (for the second term).

For \(4(3 - 2i)\):
\[
4(3 - 2i)=4\times3-4\times2i = 12 - 8i
\]

For \((2 + 3i)^2\):
Let \(a = 2\) and \(b = 3i\). Then,
\[

$$\begin{align*} (2 + 3i)^2&=2^2+2\times2\times3i+(3i)^2\\ &=4 + 12i+9i^2 \end{align*}$$

\]
Since \(i^2=- 1\), we substitute that in:
\[
4 + 12i+9\times(-1)=4 + 12i-9=-5 + 12i
\]

Step2: Add the two expanded expressions

Now, add \(12 - 8i\) and \(-5 + 12i\):
\[
(12 - 8i)+(-5 + 12i)=(12-5)+(-8i + 12i)
\]
Simplify the real and imaginary parts separately:
\[
7 + 4i
\]

Answer:

\(7 + 4i\)