%5E2019.jpeg "(1+i)^2019")
Exploring the Power of Complex Numbers: (1 + i)^2019
The expression (1 + i)^2019 might seem daunting at first glance, especially with such a large exponent. However, understanding the nature of complex numbers and their properties can simplify this calculation significantly.
De Moivre's Theorem: A Powerful Tool
The key to solving this problem lies in De Moivre's Theorem, which provides a straightforward way to calculate powers of complex numbers expressed in polar form.
De Moivre's Theorem states:
(cos θ + i sin θ)^n = cos(nθ) + i sin(nθ)
where 'n' is an integer.
Applying De Moivre's Theorem
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Convert (1 + i) to Polar Form:
- We find the magnitude (or modulus) of (1 + i) as √(1^2 + 1^2) = √2.
- The angle (or argument) can be found using the arctangent function: arctan(1/1) = π/4.
- Therefore, (1 + i) can be expressed in polar form as √2(cos(π/4) + i sin(π/4)).
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Apply De Moivre's Theorem:
- (1 + i)^2019 = [√2(cos(π/4) + i sin(π/4))]^2019
- Using De Moivre's Theorem, this simplifies to:
- √2^2019 * [cos(2019π/4) + i sin(2019π/4)]
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Simplify the Result:
- √2^2019 can be further simplified.
- 2019π/4 can be reduced to a simpler angle using the fact that the sine and cosine functions have a period of 2π.
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Final Calculation:
- After simplifying the angle and magnitude, we can express the result as a complex number in rectangular form (a + bi), where 'a' and 'b' are real numbers.
Conclusion
By employing De Moivre's Theorem, we can effectively calculate the power of a complex number like (1 + i)^2019. The process involves converting the complex number to polar form, applying the theorem, and simplifying the resulting expression. This demonstrates the power of complex number properties in solving seemingly complex problems.