(a) The complex number $z_1 = 2 + i$, where $i^2 = -1$, is shown on the Argand Diagram below - Leaving Cert Mathematics - Question 2 - 2019

The complex conjugate of $z_1$ is given by:

$\bar{z_1} = 2 - i.$
We plot $\bar{z_1}$ on the Argand Diagram at the point (2, -1) and label it accordingly.

First, we compute the magnitudes:

1. The magnitude of $z_2$ is:

$|z_2| = |4 + 2i| = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20}.$

2. We also find $z_1 + \bar{z_1}$:

$z_1 + \bar{z_1} = (2 + i) + (2 - i) = 4.$

3. Therefore, the magnitude is:

$|z_1 + \bar{z_1}| = |4| = 4.$

We need to check: $|z_2| = |z_1 + \bar{z_1}| = \sqrt{20} \neq 4.$
Thus, $|z_2| \neq |z_1 + \bar{z_1}|$.

To verify that $z_1$ is a solution of the equation, we substitute $z_1$ into the left side:

$(z_1)^2 - 4(z_1) + 5 = (2+i)^2 - 4(2+i) + 5.$

Calculating $(2+i)^2$ gives:

$(2 + i)(2 + i) = 4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i.$

Next, calculating $-4(2+i)$ gives:

$-8 - 4i.$

Now we combine all parts:

$3 + 4i - 8 - 4i + 5 = 3 - 8 + 5 = 0.$
Thus, $z_1$ is indeed a solution of the equation.