Certainly! Let's rewrite the expression [tex]\[
\left[\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)\right]^4
\][/tex]
in its standard form [tex]\(a + bi\)[/tex].
First, notice that the given complex number is in polar form:
[tex]\[
r \left( \cos \theta + i \sin \theta \right)
\][/tex]
where [tex]\( r = \sqrt{2} \)[/tex] and [tex]\( \theta = \frac{3 \pi}{4} \)[/tex].
Next, we need to convert this polar form to its equivalent rectangular form using Euler's formula:
[tex]\[
z = r \left( \cos \theta + i \sin \theta \right) = \sqrt{2} \left( \cos \frac{3 \pi}{4} + i \sin \frac{3 \pi}{4} \right)
\][/tex]
Given [tex]\( \theta = \frac{3 \pi}{4} \)[/tex]:
[tex]\[
\cos \frac{3 \pi}{4} = -\frac{\sqrt{2}}{2}
\][/tex]
[tex]\[
\sin \frac{3 \pi}{4} = \frac{\sqrt{2}}{2}
\][/tex]
So, we can write:
[tex]\[
z = \sqrt{2} \left(-\frac{\sqrt{2}}{2} + i \cdot \frac{\sqrt{2}}{2}\right)
\][/tex]
Simplify the expression inside the parentheses:
[tex]\[
z = \sqrt{2} \left(-\frac{\sqrt{2}}{2} + i \cdot \frac{\sqrt{2}}{2}\right)
= \sqrt{2} \cdot -\frac{\sqrt{2}}{2} + \sqrt{2} \cdot i \cdot \frac{\sqrt{2}}{2}
= -1 + i
\][/tex]
Now, we need to raise this result to the 4th power:
[tex]\[
(-1 + i)^4
\][/tex]
We use the Binomial Theorem to expand this:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
with [tex]\( a = -1 \)[/tex], [tex]\( b = i \)[/tex], and [tex]\( n = 4 \)[/tex]:
[tex]\[
(-1 + i)^4 = \sum_{k = 0}^{4} \binom{4}{k} (-1)^{4-k} i^k
\][/tex]
Expanding each term:
[tex]\[
\begin{align*}
k = 0: & \binom{4}{0} (-1)^4 i^0 = 1 \cdot 1 \cdot 1 = 1 \\
k = 1: & \binom{4}{1} (-1)^3 i^1 = 4 \cdot (-1) \cdot i = -4i \\
k = 2: & \binom{4}{2} (-1)^2 i^2 = 6 \cdot 1 \cdot -1 = -6 \\
k = 3: & \binom{4}{3} (-1)^1 i^3 = 4 \cdot (-1) \cdot (-i) = 4i \\
k = 4: & \binom{4}{4} (-1)^0 i^4 = 1 \cdot 1 \cdots 1 = 1
\end{align*}
\][/tex]
Adding these terms together:
[tex]\[
1 - 4i - 6 + 4i + 1 = -4
\][/tex]
Therefore, the result is:
[tex]\[
(-1 + i)^4 = -4 + 0i
\][/tex]
Summing up, the expression in the standard form [tex]\(a + bi\)[/tex] is:
[tex]\[
\boxed{-4 + 1.776356839400251e-15 i}
\][/tex]
Here the imaginary part [tex]\( 1.776356839400251 \times 10^{-15} \)[/tex] is essentially zero, so it can be treated as [tex]\(-4\)[/tex] in practice.
