Sure! Let's prove the identity [tex]\((a + bi)^2 = (a + b)(a - b) + 2abi\)[/tex] step-by-step. Here, [tex]\(i\)[/tex] represents the imaginary unit, where [tex]\(i^2 = -1\)[/tex].
### Step-by-Step Proof
1. Expanding [tex]\((a + bi)^2\)[/tex]:
[tex]\[
(a + bi)^2.
\][/tex]
Using the binomial expansion formula [tex]\((x + y)^2 = x^2 + 2xy + y^2\)[/tex],
[tex]\[
(a + bi)^2 = a^2 + 2abi + (bi)^2.
\][/tex]
Recall that [tex]\((bi)^2 = b^2i^2\)[/tex] and since [tex]\(i^2 = -1\)[/tex],
[tex]\[
(bi)^2 = b^2(-1) = -b^2.
\][/tex]
Hence,
[tex]\[
(a + bi)^2 = a^2 + 2abi - b^2.
\][/tex]
2. Expanding [tex]\((a + b)(a - b) + 2abi\)[/tex]:
[tex]\[
(a + b)(a - b) + 2abi.
\][/tex]
We can recognize that [tex]\((a + b)(a - b)\)[/tex] is a difference of squares:
[tex]\[
(a + b)(a - b) = a^2 - b^2.
\][/tex]
Hence,
[tex]\[
(a + b)(a - b) + 2abi = a^2 - b^2 + 2abi.
\][/tex]
3. Combining the Results:
[tex]\[
(a + bi)^2 = a^2 + 2abi - b^2,
\][/tex]
[tex]\[
(a + b)(a - b) + 2abi = a^2 - b^2 + 2abi.
\][/tex]
4. Comparing Both Sides:
Both expanded forms are:
[tex]\[
(a + bi)^2 = a^2 + 2abi - b^2,
\][/tex]
[tex]\[
(a + b)(a - b) + 2abi = a^2 - b^2 + 2abi.
\][/tex]
Clearly, the expressions on both sides are identical. Therefore, we have shown that:
[tex]\[
(a + bi)^2 = (a + b)(a - b) + 2abi.
\][/tex]
This completes the proof.
