What is the value of \({\left[ {\frac{{i + \sqrt 3 }}{2}} \right]^{2019}} + {\left[ {\frac{{i - \sqrt 3 }}{2}} \right]^{2019}}?\)
2i
Understanding the Complex Number Problem
The question asks us to find the value of a sum involving two complex numbers raised to a high power, 2019. The complex numbers are \({\left[ {\frac{{i + \sqrt 3 }}{2}} \right]}\) and \({\left[ {\frac{{i - \sqrt 3 }}{2}} \right]}\). To evaluate powers of complex numbers efficiently, especially high powers, it's usually best to convert them into their polar form or trigonometric form. De Moivre's theorem is then used for raising the complex number in polar form to a power.
Step-by-Step Solution using Polar Form and De Moivre's Theorem
Let's evaluate each term separately.
Term 1: Evaluating \({\left[ {\frac{{i + \sqrt 3 }}{2}} \right]^{2019}}\)
The first complex number is \(z_1 = \frac{i + \sqrt 3}{2} = \frac{\sqrt 3 + i}{2}\). We need to convert this to polar form \(r(\cos \theta + i \sin \theta)\), where \(r\) is the modulus and \(\theta\) is the argument.
- Modulus \(r_1\): \(r_1 = \left| \frac{\sqrt 3 + i}{2} \right| = \frac{1}{2} |\sqrt 3 + i| = \frac{1}{2} \sqrt{(\sqrt 3)^2 + (1)^2} = \frac{1}{2} \sqrt{3 + 1} = \frac{1}{2} \sqrt{4} = \frac{1}{2} \times 2 = 1\)
- Argument \(\theta_1\): For \(z_1 = \frac{\sqrt 3}{2} + i \frac{1}{2}\), the real part is \(\frac{\sqrt 3}{2}\) and the imaginary part is \(\frac{1}{2}\). Both are positive, so the number is in the first quadrant. \(\cos \theta_1 = \frac{\text{Real Part}}{r_1} = \frac{\sqrt 3/2}{1} = \frac{\sqrt 3}{2}\) \(\sin \theta_1 = \frac{\text{Imaginary Part}}{r_1} = \frac{1/2}{1} = \frac{1}{2}\) The angle \(\theta_1\) that satisfies these conditions is \(\frac{\pi}{6}\) radians (or 30°).
So, the polar form of \(z_1\) is \(1 \left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right)\), which can also be written in exponential form as \(e^{i\pi/6}\).
Now, we apply De Moivre's Theorem to find \(z_1^{2019}\): \(z_1^{2019} = \left(e^{i\pi/6}\right)^{2019} = e^{i \frac{2019\pi}{6}}\)
Let's simplify the angle \(\frac{2019\pi}{6}\): \(\frac{2019}{6} = 336.5\). So, \(\frac{2019\pi}{6} = 336.5\pi = \left(336 + \frac{1}{2}\right)\pi = 336\pi + \frac{\pi}{2}\).
Using the property \(e^{i(\phi + 2n\pi)} = e^{i\phi}\) for integer \(n\): \(e^{i(336\pi + \pi/2)} = e^{i\pi/2}\). Converting back to trigonometric form: \(e^{i\pi/2} = \cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right) = 0 + i(1) = i\).
So, \({\left[ {\frac{{i + \sqrt 3 }}{2}} \right]^{2019}} = i\).
Term 2: Evaluating \({\left[ {\frac{{i - \sqrt 3 }}{2}} \right]^{2019}}\)
The second complex number is \(z_2 = \frac{i - \sqrt 3}{2} = \frac{-\sqrt 3 + i}{2}\). Let's convert this to polar form.
- Modulus \(r_2\): \(r_2 = \left| \frac{-\sqrt 3 + i}{2} \right| = \frac{1}{2} |-\sqrt 3 + i| = \frac{1}{2} \sqrt{(-\sqrt 3)^2 + (1)^2} = \frac{1}{2} \sqrt{3 + 1} = \frac{1}{2} \sqrt{4} = \frac{1}{2} \times 2 = 1\)
- Argument \(\theta_2\): For \(z_2 = \frac{-\sqrt 3}{2} + i \frac{1}{2}\), the real part is \(\frac{-\sqrt 3}{2}\) and the imaginary part is \(\frac{1}{2}\). The real part is negative and the imaginary part is positive, so the number is in the second quadrant. \(\cos \theta_2 = \frac{\text{Real Part}}{r_2} = \frac{-\sqrt 3/2}{1} = -\frac{\sqrt 3}{2}\) \(\sin \theta_2 = \frac{\text{Imaginary Part}}{r_2} = \frac{1/2}{1} = \frac{1}{2}\) The angle \(\theta_2\) that satisfies these conditions in the second quadrant is \(\frac{5\pi}{6}\) radians (or 150°).
So, the polar form of \(z_2\) is \(1 \left(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6}\right)\), or \(e^{i5\pi/6}\) in exponential form.
Now, we apply De Moivre's Theorem to find \(z_2^{2019}\): \(z_2^{2019} = \left(e^{i5\pi/6}\right)^{2019} = e^{i \frac{5 \times 2019\pi}{6}} = e^{i \frac{10095\pi}{6}}\)
Let's simplify the angle \(\frac{10095\pi}{6}\): \(\frac{10095}{6} = 1682.5\). So, \(\frac{10095\pi}{6} = 1682.5\pi = \left(1682 + \frac{1}{2}\right)\pi = 1682\pi + \frac{\pi}{2}\).
Using the property \(e^{i(\phi + 2n\pi)} = e^{i\phi}\): \(e^{i(1682\pi + \pi/2)} = e^{i\pi/2}\). Converting back to trigonometric form: \(e^{i\pi/2} = \cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right) = 0 + i(1) = i\).
So, \({\left[ {\frac{{i - \sqrt 3 }}{2}} \right]^{2019}} = i\).
Calculating the Sum
The required value is the sum of the results from Term 1 and Term 2:
\({\left[ {\frac{{i + \sqrt 3 }}{2}} \right]^{2019}} + {\left[ {\frac{{i - \sqrt 3 }}{2}} \right]^{2019}} = i + i = 2i\).
Alternative Approach using Conjugates
Let \(z = \frac{\sqrt 3 + i}{2}\). We found that \(z = e^{i\pi/6}\). The first term is \(z^{2019}\). The second complex number is \(\frac{-\sqrt 3 + i}{2}\). Notice that this is the negative of the conjugate of \(z\). \(\bar{z} = \frac{\sqrt 3 - i}{2}\). \(-\bar{z} = -\left(\frac{\sqrt 3 - i}{2}\right) = \frac{-\sqrt 3 + i}{2}\). So the second term is \((-\bar{z})^{2019}\).
The expression becomes \(z^{2019} + (-\bar{z})^{2019}\). Since 2019 is an odd integer, \((-1)^{2019} = -1\). The expression simplifies to \(z^{2019} + (-1) (\bar{z})^{2019} = z^{2019} - (\bar{z})^{2019}\).
We know that \((\bar{z})^{2019} = \overline{(z^{2019})}\). Let \(w = z^{2019}\). The expression is \(w - \bar{w}\).
If \(w = a + bi\), then \(\bar{w} = a - bi\). \(w - \bar{w} = (a + bi) - (a - bi) = a + bi - a + bi = 2bi\). This means the value of the expression is 2 times the imaginary part of \(z^{2019}\), multiplied by \(i\).
From the first method, we found that \(z^{2019} = \left(\frac{\sqrt 3 + i}{2}\right)^{2019} = i\). The imaginary part of \(i\) is 1.
Therefore, the value of the expression is \(2 \times \text{Im}(i) \times i = 2 \times 1 \times i = 2i\).
Both methods yield the same result, \(2i\).
Summary of Results
| Complex Number | Polar Form | Power (2019) Calculation | Result |
|---|---|---|---|
| \(\frac{{i + \sqrt 3 }}{2}\) | \(e^{i\pi/6}\) | \(e^{i (2019\pi/6)} = e^{i(336\pi + \pi/2)} = e^{i\pi/2}\) | \(i\) |
| \(\frac{{i - \sqrt 3 }}{2}\) | \(e^{i5\pi/6}\) | \(e^{i (10095\pi/6)} = e^{i(1682\pi + \pi/2)} = e^{i\pi/2}\) | \(i\) |
Sum = \(i + i = 2i\).
Revision Table: Key Concepts
| Concept | Description | Formula/Application |
|---|---|---|
| Complex Number Polar Form | Representing \(z = x + iy\) as \(r(\cos \theta + i \sin \theta)\) or \(re^{i\theta}\) | \(r = \sqrt{x^2 + y^2}\), \(\theta = \text{atan2}(y, x)\) |
| De Moivre's Theorem | For integer \(n\), \((r(\cos \theta + i \sin \theta))^n = r^n(\cos(n\theta) + i \sin(n\theta))\) | \((re^{i\theta})^n = r^n e^{in\theta}\) |
| Argument Periodicity | Adding integer multiples of \(2\pi\) to the argument does not change the complex number | \(e^{i(\theta + 2n\pi)} = e^{i\theta}\) |
| Complex Conjugate | If \(z = x + iy\), then \(\bar{z} = x - iy\) | Properties: \(z + \bar{z} = 2 \text{Re}(z)\), \(z - \bar{z} = 2i \text{Im}(z)\), \((z^n)^- = (\bar{z})^n\) |
Additional Information: Roots of Unity Connection
The complex numbers \(\frac{\sqrt 3 + i}{2}\) and \(\frac{-\sqrt 3 + i}{2}\) are related to the 12th roots of unity. \(e^{i\pi/6}\) is a primitive 12th root of unity (\((e^{i\pi/6})^{12} = e^{i2\pi} = 1\)). \(e^{i5\pi/6}\) is also a 12th root of unity (\((e^{i5\pi/6})^{12} = e^{i10\pi} = 1\)).
Evaluating high powers of such numbers often results in one of the roots of unity or a simple value like \(\pm 1, \pm i\).
In this problem, both \({\left[ {\frac{{i + \sqrt 3 }}{2}} \right]^{2019}}\) and \({\left[ {\frac{{i - \sqrt 3 }}{2}} \right]^{2019}}\) simplified to \(e^{i\pi/2} = i\). This happened because when the arguments \(2019\pi/6\) and \(10095\pi/6\) were simplified modulo \(2\pi\), they both resulted in \(\pi/2\). \(\frac{2019\pi}{6} = \frac{673}{2}\pi = \left(336 + \frac{1}{2}\right)\pi \equiv \frac{\pi}{2} \pmod{2\pi}\) \(\frac{10095\pi}{6} = \frac{3365}{2}\pi = \left(1682 + \frac{1}{2}\right)\pi \equiv \frac{\pi}{2} \pmod{2\pi}\)
The sum \(i + i\) then gives the final answer \(2i\).
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