z = \frac{4}{1 + \sqrt{3}i} is a complex number, where \: i^2 = -1 - Leaving Cert Mathematics - Question 1 - 2013

To plot z on the Argand diagram, we first need to identify the real and imaginary parts of z:

The expression for z is 1 - \sqrt{3}i, where the real part is 1 and the imaginary part is -\sqrt{3}. Thus, we can plot the point (1, -\sqrt{3}) on the Argand diagram, located in the fourth quadrant.

Next, we convert z to polar form:

1. Calculate the modulus (r): = \sqrt{1 + 3} = \sqrt{4} = 2.$$
2. Find the argument (\theta):
   $\tan(\alpha) = \frac{-\sqrt{3}}{1} \Rightarrow \alpha = -\frac{\pi}{3}.$
   This gives us the polar form:
   $z = r(e^{i\theta}) = 2(cos(-\frac{\pi}{3}) + i sin(-\frac{\pi}{3})) = 2(cos(\frac{5\pi}{3}) + i sin(\frac{5\pi}{3})).$

Using De Moivre's theorem:

= 2^{10}\left(cos\left(10 \cdot \frac{5\pi}{3}\right) + isin\left(10 \cdot \frac{5\pi}{3}\right)\right) = 2^{10}\left(cos\left(\frac{50\pi}{3}\right) + isin\left(\frac{50\pi}{3}\right)\right) = 2^{10}\left(cos(\frac{50\pi}{3} - 2\pi \cdot 8) + isin(\frac{50\pi}{3} - 2\pi \cdot 8)\right) = 2^{10}\left(cos(\frac{2\pi}{3}) + isin(\frac{2\pi}{3})\right) = 2^{10}\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -2^{10}(1 - \sqrt{3}i).$$