The complex number z has modulus \( \frac{5}{3} \) and argument \( \frac{4\pi}{9} \) - Leaving Cert Mathematics - Question 3 - 2012

To find the fourth roots, we use the formula for the roots of a complex number:

[ w_n = r^{1/4} \left( \cos \left( \frac{\theta + 2\pi n}{4} \right) + i \sin \left( \frac{\theta + 2\pi n}{4} \right) \right) ]

Here, ( r^{1/4} = \left( \frac{5}{3} \right)^{1/4} = \frac{5^{1/4}}{3^{1/4}} ), and ( \theta = \frac{4\pi}{9} ).

We proceed by calculating the four values for ( n = 0, 1, 2, 3 ):

1. ( n = 0 ): [ w_0 = \frac{5^{1/4}}{3^{1/4}} \left( \cos \left( \frac{4\pi}{36} \right) + i \sin \left( \frac{4\pi}{36} \right) \right) = \frac{5^{1/4}}{3^{1/4}} \left( \cos \left( \frac{\pi}{9} \right) + i \sin \left( \frac{\pi}{9} \right) \right) ]

2. ( n = 1 ): [ w_1 = \frac{5^{1/4}}{3^{1/4}} \left( \cos \left( \frac{10\pi}{36} \right) + i \sin \left( \frac{10\pi}{36} \right) \right) = \frac{5^{1/4}}{3^{1/4}} \left( \cos \left( \frac{5\pi}{18} \right) + i \sin \left( \frac{5\pi}{18} \right) \right) ]

3. ( n = 2 ): [ w_2 = \frac{5^{1/4}}{3^{1/4}} \left( \cos \left( \frac{16\pi}{36} \right) + i \sin \left( \frac{16\pi}{36} \right) \right) = \frac{5^{1/4}}{3^{1/4}} \left( \cos \left( \frac{8\pi}{9} \right) + i \sin \left( \frac{8\pi}{9} \right) \right) ]

4. ( n = 3 ): [ w_3 = \frac{5^{1/4}}{3^{1/4}} \left( \cos \left( \frac{22\pi}{36} \right) + i \sin \left( \frac{22\pi}{36} \right) \right) = \frac{5^{1/4}}{3^{1/4}} \left( \cos \left( \frac{11\pi}{18} \right) + i \sin \left( \frac{11\pi}{18} \right) \right) ]

On the Argand diagram, each of the computed roots ( w_0, w_1, w_2, w_3 ) should be plotted based on their real and imaginary components. The positions will represent the angle and modulus of each root, visualizing their distribution around the origin in the complex plane.