(a) Write each of the following complex numbers in the form $a + bi$, where $i^2 = -1$ - Leaving Cert Mathematics - Question 3 - 2013

Calculating the first part:
$z_2 = (5 + 4i)(17 - 13i)$

Using the distributive property:
$= 85 - 65i + 68i - 52i^2$
$= 85 + 52 + 3i = 137 + 3i$.

Now, calculating the second part:
$= (5 + 3i)(17 - 13i)$
= $85 - 65i + 51i - 39i^2$
$= 85 + 39 + (-14)i = 124 - 14i$.

Subtracting gives:
$z_2 = (137 + 3i) - (124 - 14i) = 13 + 17i$.

For the first term:
$\left( \frac{5}{2} + \frac{7}{2} i \right)^2 = \frac{25}{4} + 2 \cdot (\frac{5}{2})(\frac{7}{2}) i + (\frac{7}{2} i)^2 = \frac{25}{4} + \frac{35}{2} i - \frac{49}{4} = -\frac{24}{4} + \frac{35}{2} i = -6 + \frac{35}{2} i.$

For the second term:
$\left( \frac{5}{2} - \frac{1}{2} i \right)^2 = \frac{25}{4} - 2 \cdot (\frac{5}{2})(\frac{1}{2}) i + (\frac{1}{2} i)^2 = \frac{25}{4} - \frac{5}{2} i - \frac{1}{4} = \frac{24}{4} - \frac{5}{2} i = 6 - \frac{5}{2} i.$

Now, we subtract the two results to find $z_3$:
$z_3 = (-6 + \frac{35}{2} i) - (6 - \frac{5}{2} i) = -12 + 20i.$

Breaking this down:
Recall that $i^2 = -1$ and $i^3 = -i$,
Therefore:
$z_4 = 1 + i - 1 - i = 0 + 0i = 0.$

To determine which complex number is farther from the origin, we calculate the modulus of each.
For $z_1 = 16 - 11i$, the modulus is:
$|z_1| = \sqrt{(16)^2 + (-11)^2} = \sqrt{256 + 121} = \sqrt{377}$.

For $z_2 = 13 + 17i$, the modulus is:
$|z_2| = \sqrt{(13)^2 + (17)^2} = \sqrt{169 + 289} = \sqrt{458}$.

Thus, $|z_2| > |z_1|$, indicating that $z_2$ is farther from 0 on the Argand diagram.