Finding the Argument of a Complex Number

The problem asks us to find the argument of the complex number \(z = \frac{1+i √{3}}{1-i √{3}}\), where \(i = √{-1}\).

We can solve this problem using two main methods:

1. Using the property of arguments for division of complex numbers.
2. Simplifying the complex number \(z\) first and then finding its argument.

Method 1: Using Argument Property

Let the numerator be \(z_1 = 1 + i\sqrt{3}\) and the denominator be \(z_2 = 1 - i\sqrt{3}\). Then \(z = \frac{z_1}{z_2}\).

The argument of a quotient of complex numbers is given by:

\(\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)\)

We need to find the arguments of \(z_1\) and \(z_2\).

Argument of \(z_1 = 1 + i\sqrt{3}\)

\(z_1\) is in the form \(x + iy\), where \(x = 1\) and \(y = \sqrt{3}\).

Since \(x > 0\) and \(y > 0\), \(z_1\) lies in the first quadrant. The argument (principal value) is given by:

\(\arg(z_1) = \tan^{-1}\left(\frac{y}{x}\right)\)

\(\arg(z_1) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \tan^{-1}(\sqrt{3})\)

We know that \(\tan(\frac{\pi}{3}) = \sqrt{3}\).

So, \(\arg(z_1) = \frac{\pi}{3}\).

Argument of \(z_2 = 1 - i\sqrt{3}\)

\(z_2\) is in the form \(x + iy\), where \(x = 1\) and \(y = -\sqrt{3}\).

Since \(x > 0\) and \(y < 0\), \(z_2\) lies in the fourth quadrant. The argument (principal value) is given by:

\(\arg(z_2) = \tan^{-1}\left(\frac{y}{x}\right)\)

\(\arg(z_2) = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = \tan^{-1}(-\sqrt{3})\)

In the fourth quadrant, the principal argument is negative. Since \(\tan(-\theta) = -\tan(\theta)\), and \(\tan(\frac{\pi}{3}) = \sqrt{3}\), we have \(\tan(-\frac{\pi}{3}) = -\sqrt{3}\).

So, \(\arg(z_2) = -\frac{\pi}{3}\).

Calculating Argument of \(z\)

Now, we can find \(\arg(z)\):

\(\arg(z) = \arg(z_1) - \arg(z_2)\)

\(\arg(z) = \frac{\pi}{3} - \left(-\frac{\pi}{3}\right)\)

\(\arg(z) = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}\)

Method 2: Simplifying \(z\) First

We can simplify the complex number \(z\) by multiplying the numerator and denominator by the conjugate of the denominator.

The denominator is \(1 - i\sqrt{3}\). Its conjugate is \(1 + i\sqrt{3}\).

\(z = \frac{1+i\sqrt{3}}{1-i\sqrt{3}} \times \frac{1+i\sqrt{3}}{1+i\sqrt{3}}\)

Numerator: \((1+i\sqrt{3})(1+i\sqrt{3}) = (1)^2 + 2(1)(i\sqrt{3}) + (i\sqrt{3})^2 = 1 + 2i\sqrt{3} + i^2(3) = 1 + 2i\sqrt{3} - 3 = -2 + 2i\sqrt{3}\)

Denominator: \((1-i\sqrt{3})(1+i\sqrt{3}) = (1)^2 - (i\sqrt{3})^2 = 1 - i^2(3) = 1 - (-3) = 1 + 3 = 4\)

So, \(z = \frac{-2 + 2i\sqrt{3}}{4} = \frac{-2}{4} + \frac{2i\sqrt{3}}{4} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\).

Argument of \(z = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\)

This simplified \(z\) is in the form \(x + iy\), where \(x = -\frac{1}{2}\) and \(y = \frac{\sqrt{3}}{2}\).

Since \(x < 0\) and \(y > 0\), \(z\) lies in the second quadrant.

To find the principal argument in the second quadrant, we first find the reference angle \(\alpha = \tan^{-1}\left|\frac{y}{x}\right|\).

\(\alpha = \tan^{-1}\left|\frac{\sqrt{3}/2}{-1/2}\right| = \tan^{-1}|-\sqrt{3}| = \tan^{-1}(\sqrt{3})\)

We know that \(\tan(\frac{\pi}{3}) = \sqrt{3}\), so the reference angle is \(\alpha = \frac{\pi}{3}\).

For a complex number in the second quadrant, the principal argument is \(\pi - \alpha\).

\(\arg(z) = \pi - \frac{\pi}{3} = \frac{3\pi - \pi}{3} = \frac{2\pi}{3}\)

Conclusion

Both methods yield the same result for the argument of \(z\).

The argument of \(z = \frac{1+i √{3}}{1-i √{3}}\) is \(\frac{2\pi}{3}\).

Revision Table: Complex Number Arguments

Additional Information on Complex Numbers and Arguments

A complex number \(z = x + iy\) can be represented in the complex plane as a point \((x, y)\).

• The modulus or magnitude of \(z\) is the distance from the origin to the point \((x, y)\), denoted by \(|z|\) or \(r\), calculated as \(r = \sqrt{x^2 + y^2}\).
• The argument of \(z\), denoted by \(\arg(z)\) or \(\theta\), is the angle between the positive x-axis and the line segment connecting the origin to the point \((x, y)\). The principal argument is usually restricted to the interval \((-\pi, \pi]\).

The polar form of a complex number \(z\) is given by \(z = r(\cos \theta + i \sin \theta)\).

When dividing two complex numbers in polar form, \(z_1 = r_1(\cos \theta_1 + i \sin \theta_1)\) and \(z_2 = r_2(\cos \theta_2 + i \sin \theta_2)\), their quotient is:

\(\frac{z_1}{z_2} = \frac{r_1}{r_2}(\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2))\)

From this polar form of the quotient, we can see that the modulus is the ratio of the moduli (\(r_1/r_2\)), and the argument is the difference of the arguments (\(\theta_1 - \theta_2\)), which confirms the property used in Method 1.