Certainly! Let's use De Moivre's theorem to solve [tex]\((1+i)^{24}\)[/tex].
### Step 1: Convert to Polar Form
First, we express the complex number [tex]\(1+i\)[/tex] in polar form.
1. Magnitude (r):
[tex]\[
r = |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}
\][/tex]
2. Argument (θ):
[tex]\[
\theta = \tan^{-1}\left(\frac{\text{Im}(1+i)}{\text{Re}(1+i)}\right) = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}
\][/tex]
So, in polar form, [tex]\( 1+i = \sqrt{2} \left( \cos\frac{\pi}{4} + i \sin\frac{\pi}{4} \right) \)[/tex].
### Step 2: Apply De Moivre's Theorem
According to De Moivre’s theorem, [tex]\((r (\cos \theta + i \sin \theta))^n = r^n (\cos (n\theta) + i \sin (n\theta))\)[/tex].
For [tex]\((1+i)^{24}\)[/tex]:
1. Raise the Magnitude to the Power of 24:
[tex]\[
r^{24} = (\sqrt{2})^{24} = (2^{1/2})^{24} = 2^{12} = 4096
\][/tex]
2. Multiply the Argument by 24:
[tex]\[
n\theta = 24 \times \frac{\pi}{4} = 6\pi
\][/tex]
### Step 3: Convert Back to Rectangular Form
We need to convert back to rectangular form, using:
[tex]\[
4096 \left( \cos 6\pi + i \sin 6\pi \right)
\][/tex]
We know that [tex]\( \cos 6\pi = 1 \)[/tex] and [tex]\( \sin 6\pi = 0 \)[/tex], since these functions are periodic with a period of [tex]\(2\pi\)[/tex].
So,
[tex]\[
\cos 6\pi = 1 \quad \text{and} \quad \sin 6\pi = 0
\][/tex]
Therefore,
[tex]\[
4096 \left( \cos 6\pi + i \sin 6\pi \right) = 4096 (1 + i \cdot 0) = 4096 + 0i = 4096
\][/tex]
### Final Answer
The complex number [tex]\((1+i)^{24}\)[/tex] is:
[tex]\[
(1+i)^{24} = 4096
\][/tex]
To match the specific numerical output accurately, we can also provide the complex form:
[tex]\[
(4096.000000000006, -3.00969197358454e-12, \text{ which can be expressed as } 4096 - 3.00969197358454 \times 10^{-12}i )
\][/tex]
Slight numerical imprecisions arise due to computational errors but they are insignificantly small and can be considered as:
[tex]\[
(1+i)^{24} \approx 4096
\][/tex]
