jee 2010 paper.....let omega be the complex number cos(2pi/3)+isin(2pi/3). Then the number of distinct complex numbers z satisfying

Let's delve into the problem involving the complex number \(\omega\) defined as \(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\). This expression can also be represented in exponential form using Euler's formula: \(\omega = e^{i\frac{2\pi}{3}}\). To find the number of distinct complex numbers \(z\) that satisfy a given condition involving \(\omega\), we first need to clarify the equation or inequality these \(z\) must satisfy. For the sake of explanation, let’s assume we are looking for the solutions to an equation of the form:\[z^n = \omega\]where \(n\) is some positive integer.

Understanding Roots of Complex Numbers

From complex number theory, we know that the solutions to equations of the form \(z^n = w\) (where \(w\) is a complex number) can be found using the concept of roots of unity. Specifically, the \(n\)th roots of a complex number are given by:\[z_k = |w|^{1/n} \left( \cos\left( \frac{\arg(w) + 2k\pi}{n} \right) + i\sin\left( \frac{\arg(w) + 2k\pi}{n} \right) \right)\]for \(k = 0, 1, 2, \ldots, n-1\).

Application to Our Problem

In our case, \(\omega\) has a modulus (or absolute value) of 1 since both \(\cos\left(\frac{2\pi}{3}\right)\) and \(\sin\left(\frac{2\pi}{3}\right)\) yield coordinates on the unit circle. The argument of \(\omega\) is \(\frac{2\pi}{3}\). Therefore, the \(n\)th roots of \(\omega\) can be expressed as:

• Magnitude: \(1^{1/n} = 1\)
• Argument: \(\frac{\frac{2\pi}{3} + 2k\pi}{n}\) for \(k = 0, 1, 2, \ldots, n-1\)

This gives us the distinct complex solutions \(z_k\):\[z_k = \cos\left(\frac{\frac{2\pi}{3} + 2k\pi}{n}\right) + i\sin\left(\frac{\frac{2\pi}{3} + 2k\pi}{n}\right)\]for each integer \(k\) in the specified range.

Distinct Solutions

Now, since \(k\) takes on values from \(0\) to \(n-1\), we find that there will be \(n\) distinct complex numbers \(z_k\) that satisfy the equation \(z^n = \omega\). Each of these will correspond to a unique angle on the unit circle, spaced evenly apart by \(\frac{2\pi}{n}\).

Example Illustration

Let’s take an example where \(n = 3\). The three distinct complex numbers \(z_k\) satisfying \(z^3 = \omega\) can be found as follows:

• For \(k = 0\):\[z_0 = \cos\left(\frac{2\pi/3 + 0}{3}\right) + i\sin\left(\frac{2\pi/3 + 0}{3}\right) = \cos\left(\frac{2\pi}{9}\right) + i\sin\left(\frac{2\pi}{9}\right)\
• For \(k = 1\):\[z_1 = \cos\left(\frac{2\pi/3 + 2\pi}{3}\right) + i\sin\left(\frac{2\pi/3 + 2\pi}{3}\right) = \cos\left(\frac{8\pi}{9}\right) + i\sin\left(\frac{8\pi}{9}\right)\
• For \(k = 2\):\[z_2 = \cos\left(\frac{2\pi/3 + 4\pi}{3}\right) + i\sin\left(\frac{2\pi/3 + 4\pi}{3}\right) = \cos\left(\frac{14\pi}{9}\right) + i\sin\left(\frac{14\pi}{9}\right)\

These calculations illustrate how the roots distribute evenly around the unit circle, confirming that there are indeed \(n\) distinct solutions.

Final Thoughts

In summary, the number of distinct complex numbers \(z\) satisfying the equation \(z^n = \omega\) is directly tied to the value of \(n\). For any integer \(n\), you will find \(n\) distinct solutions, each representing a unique point on the unit circle corresponding to that root of the complex number \(\omega\).