Understanding the Complex Number Problem

The problem asks us to find the imaginary part of a complex number \(z\), which is defined as the sum of two complex numbers raised to the power of 107.

The given expression for \(z\) is:

\(z = {\left( {\frac{{\sqrt 3 }}{2} + \frac{{\rm{i}}}{2}} \right)^{107}} + {\left( {\frac{{\sqrt 3 }}{2} - \frac{{\rm{i}}}{2}} \right)^{107}}\)

Let's analyze the two complex numbers involved:

• The first complex number is \(w_1 = \frac{{\sqrt 3 }}{2} + \frac{{\rm{i}}}{2}\).
• The second complex number is \(w_2 = \frac{{\sqrt 3 }}{2} - \frac{{\rm{i}}}{2}\).

Notice that \(w_2\) is the conjugate of \(w_1\), i.e., \(w_2 = \overline{w_1}\).

Using the Property of Complex Conjugates

We can rewrite the expression for \(z\) using this observation. Let \(w = \frac{{\sqrt 3 }}{2} + \frac{{\rm{i}}}{2}\). Then the expression becomes:

\(z = w^{107} + (\overline{w})^{107}\)

A useful property of complex numbers is that the conjugate of a power is the power of the conjugate: \((\overline{w})^n = \overline{(w^n)}\). Using this property, we get:

\(z = w^{107} + \overline{(w^{107})}\)

Let \(W = w^{107}\). Then \(z = W + \overline{W}\).

Consider any complex number \(W = a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. Its conjugate is \(\overline{W} = a - bi\).

The sum of \(W\) and \(\overline{W}\) is:

\(W + \overline{W} = (a + bi) + (a - bi) = a + bi + a - bi = 2a\)

The result \(2a\) is a purely real number. A purely real number has an imaginary part equal to zero.

Therefore, \(z\), being the sum of \(W = w^{107}\) and its conjugate \(\overline{W} = \overline{(w^{107})}\), must be a purely real number.

The imaginary part of a purely real number is always 0.

Alternative Method: Using Polar Form and De Moivre's Theorem

We can also solve this by converting the complex numbers into polar form and using De Moivre's Theorem.

Let's convert \(w_1 = \frac{{\sqrt 3 }}{2} + \frac{{\rm{i}}}{2}\) to polar form \(r(\cos \theta + i \sin \theta)\).

• Modulus \(r = \left| \frac{{\sqrt 3 }}{2} + \frac{{\rm{i}}}{2} \right| = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1\).
• Argument \(\theta = \arctan\left(\frac{1/2}{\sqrt{3}/2}\right) = \arctan\left(\frac{1}{\sqrt{3}}\right)\). Since the complex number is in the first quadrant, \(\theta = \frac{\pi}{6}\).

So, \(w_1 = 1 \left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right)\).

Now let's convert \(w_2 = \frac{{\sqrt 3 }}{2} - \frac{{\rm{i}}}{2}\) to polar form \(r(\cos \phi + i \sin \phi)\).

• Modulus \(r = \left| \frac{{\sqrt 3 }}{2} - \frac{{\rm{i}}}{2} \right| = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1\).
• Argument \(\phi = \arctan\left(\frac{-1/2}{\sqrt{3}/2}\right) = \arctan\left(-\frac{1}{\sqrt{3}}\right)\). Since the complex number is in the fourth quadrant, \(\phi = -\frac{\pi}{6}\) or \(\frac{11\pi}{6}\). Let's use \(-\frac{\pi}{6}\).

So, \(w_2 = 1 \left(\cos \left(-\frac{\pi}{6}\right) + i \sin \left(-\frac{\pi}{6}\right)\right)\).

Now, we use De Moivre's Theorem, which states that for a complex number \(r(\cos \theta + i \sin \theta)\) and an integer \(n\), \((r(\cos \theta + i \sin \theta))^n = r^n(\cos n\theta + i \sin n\theta)\).

First term:

\({\left( {\frac{{\sqrt 3 }}{2} + \frac{{\rm{i}}}{2}} \right)^{107}} = \left(1 \left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right)\right)^{107} = 1^{107} \left(\cos \left(107 \times \frac{\pi}{6}\right) + i \sin \left(107 \times \frac{\pi}{6}\right)\right)

\(= \cos \frac{107\pi}{6} + i \sin \frac{107\pi}{6}\)

Let's simplify the angle \(\frac{107\pi}{6}\). We can write \(\frac{107}{6} = \frac{102+5}{6} = 17 + \frac{5}{6}\). So \(\frac{107\pi}{6} = 17\pi + \frac{5\pi}{6}\).

Using the properties of trigonometric functions:

• \(\cos(17\pi + x) = \cos(\pi + x) = -\cos x\) (since 17 is odd)
• \(\sin(17\pi + x) = \sin(\pi + x) = -\sin x\) (since 17 is odd)

So, \(\cos \frac{107\pi}{6} = \cos \left(17\pi + \frac{5\pi}{6}\right) = -\cos \frac{5\pi}{6}\) and \(\sin \frac{107\pi}{6} = \sin \left(17\pi + \frac{5\pi}{6}\right) = -\sin \frac{5\pi}{6}\).

The first term is \(= -\cos \frac{5\pi}{6} - i \sin \frac{5\pi}{6}\).

Second term:

\({\left( {\frac{{\sqrt 3 }}{2} - \frac{{\rm{i}}}{2}} \right)^{107}} = \left(1 \left(\cos \left(-\frac{\pi}{6}\right) + i \sin \left(-\frac{\pi}{6}\right)\right)\right)^{107} = 1^{107} \left(\cos \left(107 \times -\frac{\pi}{6}\right) + i \sin \left(107 \times -\frac{\pi}{6}\right)\right)

\(= \cos \left(-\frac{107\pi}{6}\right) + i \sin \left(-\frac{107\pi}{6}\right)\)

Using the properties \(\cos(-x) = \cos x\) and \(\sin(-x) = -\sin x\):

\(= \cos \left(\frac{107\pi}{6}\right) - i \sin \left(\frac{107\pi}{6}\right)\)

From the first term calculation, we know \(\cos \frac{107\pi}{6} = -\cos \frac{5\pi}{6}\) and \(\sin \frac{107\pi}{6} = -\sin \frac{5\pi}{6}\).

So, the second term is \(= (-\cos \frac{5\pi}{6}) - i (-\sin \frac{5\pi}{6}) = -\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6}\).

Now, let's add the two terms to find \(z\):

\(z = \left(-\cos \frac{5\pi}{6} - i \sin \frac{5\pi}{6}\right) + \left(-\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6}\right)\)

\(z = -\cos \frac{5\pi}{6} - \cos \frac{5\pi}{6} - i \sin \frac{5\pi}{6} + i \sin \frac{5\pi}{6}\)

\(z = -2 \cos \frac{5\pi}{6} + ( - i \sin \frac{5\pi}{6} + i \sin \frac{5\pi}{6} ) \)

\(z = -2 \cos \frac{5\pi}{6} + 0i\)

The imaginary part of \(z\) is the coefficient of \(i\), which is 0.

Both methods confirm that the imaginary part of \(z\) is 0.

Final Answer

The imaginary part of \(z\) is 0.

Revision Table: Key Concepts

Additional Information: Understanding Complex Number Powers

Calculating high powers of complex numbers in rectangular form (\(a+bi\)) is usually very tedious. Converting the complex number to polar form is the standard approach, especially when the modulus is 1.

For a complex number \(w = r(\cos \theta + i \sin \theta) = re^{i\theta}\), its power \(w^n\) is \(r^n(\cos n\theta + i \sin n\theta) = r^n e^{in\theta}\). When \(r=1\), this simplifies to \(w^n = \cos n\theta + i \sin n\theta = e^{in\theta}\).

In this problem, the base complex numbers \(\frac{{\sqrt 3 }}{2} + \frac{{\rm{i}}}{2}\) and \(\frac{{\sqrt 3 }}{2} - \frac{{\rm{i}}}{2}\) both have a modulus of 1. This is a key indicator that polar form and De Moivre's theorem will be useful, or that properties related to roots of unity (which have modulus 1) might apply, although in this specific case, the conjugate property is more direct.

The angle \(\frac{\pi}{6}\) corresponds to 30 degrees. The values \(\cos(\pi/6) = \frac{\sqrt{3}}{2}\) and \(\sin(\pi/6) = \frac{1}{2}\) are common angles often appearing in such problems.