Given: $z_1 = -2 + 3i$ and $z_2 = -3 - 2i$, where $i^2 = -1$ - Leaving Cert Mathematics - Question 2 - 2016

1. Plot $z_1 = -2 + 3i$: Position this point at coordinates (-2, 3).
2. Plot $z_2 = -3 - 2i$: Position this point at coordinates (-3, -2).
3. Plot $z_3 = 1 + 5i$: Position this point at coordinates (1, 5).

Ensure that all points are correctly labeled.

1. Calculate $|z_1|$: $|z_1| = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$

2. Calculate $|z_2|$: $|z_2| = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$

3. Calculate $|z_3|$: $|z_3| = \sqrt{(1)^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26}$

4. Check the equality: $|z_3| = \sqrt{26} \neq \sqrt{13} + \sqrt{13} = 2\sqrt{13}$

Conclusion: $|z_3| \neq |z_1| + |z_2|$.

1. Substituting in values: $z_4 = \frac{-2 + 3i}{-3 - 2i}$

2. Multiply by the conjugate of the denominator: $z_4 = \frac{(-2 + 3i)(-3 + 2i)}{(-3 - 2i)(-3 + 2i)}$ Simplifying:

   • The denominator becomes: $(-3)^2 - (2i)^2 = 9 + 4 = 13$
   • The numerator: $(-2)(-3) + (-2)(2i) + (3i)(-3) + (3i)(2i) = 6 - 4i - 9i - 6 = 6 - 13i$

3. Final expression: $z_4 = \frac{6 - 13i}{13} = \frac{6}{13} - i\frac{13}{13} = \frac{6}{13} - i$