Understanding De Moivre's Theorem
De Moivre's Theorem is a powerful tool in complex analysis that connects complex numbers and trigonometry. It states that for any real number $\theta$ and any integer $n$:
$$(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)$$
This theorem is incredibly useful because it simplifies raising complex numbers to high powers, turning an otherwise tedious algebraic expansion into a straightforward multiplication problem.
Solving for (1 + i)^20
To apply the theorem to $(1 + i)^{20}$, we first need to represent the complex number $z = 1 + i$ in its polar (or trigonometric) form, which is $z = r(\cos \theta + i \sin \theta)$.
Step 1: Find the Modulus (r)
The modulus $r$ is the magnitude of the complex number: $$r = \sqrt{a^2 + b^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$$
Step 2: Find the Argument (θ)
The argument $\theta$ is the angle formed with the positive real axis: $$\tan \theta = \frac{b}{a} = \frac{1}{1} = 1$$ Since $(1, 1)$ lies in the first quadrant, $\theta = \frac{\pi}{4}$.
Step 3: Write in Polar Form
Now we can express $1 + i$ as: $$1 + i = \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$$
Step 4: Apply De Moivre's Theorem
Now, raise the polar form to the power of 20: $$(1 + i)^{20} = [\sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})]^{20}$$ $$(1 + i)^{20} = (\sqrt{2})^{20} \cdot [\cos(20 \cdot \frac{\pi}{4}) + i \sin(20 \cdot \frac{\pi}{4})]$$
Step 5: Simplify the Result
- Calculate the power of the modulus: $(\sqrt{2})^{20} = 2^{1/2 \cdot 20} = 2^{10} = 1024$.
- Calculate the angles: $20 \cdot \frac{\pi}{4} = 5\pi$.
- Evaluate the trigonometric functions:
- $\cos(5\pi) = \cos(\pi) = -1$
- $\sin(5\pi) = \sin(\pi) = 0$
Plugging these back into the equation: $$(1 + i)^{20} = 1024(-1 + i \cdot 0)$$ $$(1 + i)^{20} = -1024$$
Conclusion
By converting to polar form, we transformed a complex algebraic power calculation into a simple sequence of exponentiation and trigonometric evaluation. Using De Moivre's theorem, we found that $(1 + i)^{20} = -1024$.