Let $z_1 = 5 - i$ and $z_2 = 4 + 3i$, where $i^2 = -1$ - Leaving Cert Mathematics - Question 2 - 2014

$|z_1 - z_2| = |1 - 4i| = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17}.$

Similarly,

$|z_2 - z_1| = |(4 + 3i) - (5 - i)| = |(4 - 5) + (3 + 1)i| = |-1 + 4i| = \sqrt{(-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}.$

Thus, $|z_1 - z_2| = |z_2 - z_1|$.

The absolute values of the differences $z - w$ and $w - z$ are equal because they represent the same distance in the complex plane. For any two complex numbers, the distance from $z$ to $w$ is the same as the distance from $w$ to $z$. Thus, we have $|z - w| = |w - z|$.

Given $z_1 = z_2/z_3$, we rearrange this to find $z_3$. Thus, $z_3 = \frac{z_2}{z_1} = \frac{4 + 3i}{5 - i}$. Multiplying the numerator and denominator by the conjugate of the denominator:

$z_3 = \frac{(4 + 3i)(5 + i)}{(5 - i)(5 + i)} = \frac{(20 + 4i + 15i + 3i^2)}{25 + 1} = \frac{(20 + 19i - 3)}{26} = \frac{17 + 19i}{26} = \frac{17}{26} + \frac{19}{26} i.$

Hence, $z_3 = \frac{17}{26} + \frac{19}{26} i$.