z_1 = 3 - 4i, z_2 = -2 + i and z_3 = 2i z_2, where i^2 = -1 - Leaving Cert Mathematics - Question 3 - 2018

The points can be plotted as follows:

• $z_1 = 3 - 4i$ is at (3, -4).
• $z_2 = -2 + i$ is at (-2, 1).
• $z_3 = -2 - 4i$ is at (-2, -4). Label each point clearly on the Argand diagram.

The modulus of a complex number $z_2 = a + bi$ is given by:

$$|z_2| = ext{sqrt}(a^2 + b^2).$$

Here, $a = -2$ and $b = 1$. Thus:

$$|z_2| = ext{sqrt}((-2)^2 + (1)^2) = ext{sqrt}(4 + 1) = ext{sqrt}(5).$$

Given $z_1 x z_4 = 29 + 3i$, we can solve for $z_4$:

First, calculate $z_1$:

$$z_1 = 3 - 4i.$$

Now, we can rearrange the equation:

$$z_4 = \frac{29 + 3i}{z_1} = \frac{29 + 3i}{3 - 4i}.$$

Multiply the numerator and the denominator by the conjugate of the denominator:

$$z_4 = \frac{(29 + 3i)(3 + 4i)}{(3 - 4i)(3 + 4i)} = \frac{(87 + 116i + 9i + 12(-1))}{9 + 16} = \frac{(75 + 125i)}{25}.$$

Thus:

z_4 = 3 + 5i.$