(4 - 2i) = 0 + ki, where k ∈ ℤ, and $i^2 = -1$ - Leaving Cert Mathematics - Question 1 - 2021

To find the square root of $-5 + 12i$, we start with the expression:

$\sqrt{r(cos(\theta) + i sin(\theta))}$.
First, we determine the modulus, r:

$r = \sqrt{(-5)^2 + (12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13.$
Next, we need to find the argument, (\theta):

$\theta = \tan^{-1}\left(\frac{12}{-5}\right) = \tan^{-1}(-\frac{12}{5}) + \pi$.
Computing the argument gives us $\theta \approx 2.158$.
Now we can express this in polar form:

$\sqrt{13(cos(\theta) + i sin(\theta))} = \sqrt{13}(cos(\frac{\theta}{2}) + i sin(\frac{\theta}{2}))$.
Calculating gives us:

You will find two roots: $\pm (3 + 2i)$.

To find the roots of $z^3 = -8$, we first express -8 in polar form:

$-8 = 8(cos(\pi) + i sin(\pi))$.
Using De Moivre's theorem, we can find the roots:

$z = \sqrt[3]{8}(cos(\frac{\pi + 2k\pi}{3}) + i sin(\frac{\pi + 2k\pi}{3}))$, where k = 0, 1, 2.
This results in:

For $k = 0$:
$z_0 = 2(cos(\frac{\pi}{3}) + i sin(\frac{\pi}{3})) \rightarrow z_0 = 1 + \sqrt{3}i$.
For $k = 1$:
$z_1 = 2(cos(\pi) + i sin(\pi)) \rightarrow z_1 = -2$.
For $k = 2$:
$z_2 = 2(cos(\frac{5\pi}{3}) + i sin(\frac{5\pi}{3})) \rightarrow z_2 = 1 - \sqrt{3}i.$
Thus, the roots are: $1 + \sqrt{3}i, -2, 1 - \sqrt{3}i$.