1.3 Q-1 | Complex Numbers | Mathematics 11

Exercise 1.3 — Question 1

This exercise covers two key skills:

Factorizing expressions of the form $z^{2} + a^{2}$ into linear factors over $C$
Solving quadratic equations by completing the square

Over the real numbers, $z^{2} + a^{2}$ cannot be factorized. However, over $C$ , we use $i^{2} = - 1$ to rewrite it as a difference of squares:

$z^{2} + a^{2} = z^{2} - (ai)^{2} = (z + ai) (z - ai)$

Example: Factorize $3 z^{2} + 363$

Step 1: Factor out the common factor: $3 z^{2} + 363 = 3 (z^{2} + 121)$

Step 2: Apply the sum-of-squares factorization with $a = 11$ : $z^{2} + 121 = z^{2} - (11 i)^{2} = (z + 11 i) (z - 11 i)$

Result: $3 z^{2} + 363 = 3 (z + 11 i) (z - 11 i)$

To solve $p z^{2} + q z + r = 0$ by completing the square:

Step 1: Divide through by $p$ (if $p \neq = 1$ ): $z^{2} + \frac{q}{p} z + \frac{r}{p} = 0$

Step 2: Move the constant to the right: $z^{2} + \frac{q}{p} z = - \frac{r}{p}$

Step 3: Add $(\frac{q}{2 p})^{2}$ to both sides: $z^{2} + \frac{q}{p} z + (\frac{q}{2 p})^{2} = (\frac{q}{2 p})^{2} - \frac{r}{p}$

Step 4: Write the left side as a perfect square: $(z + \frac{q}{2 p})^{2} = \frac{q ^{2} - 4 p r}{4 p ^{2}}$

Step 5: Take the square root of both sides and solve for $z$ .

Example: Solve $z^{2} + 6 z + 13 = 0$

$z^{2} + 6 z = - 13$ $z^{2} + 6 z + 9 = - 13 + 9$ $(z + 3)^{2} = - 4$ $z + 3 = \pm - 4 = \pm 2 i$ $z = - 3 + 2 i or z = - 3 - 2 i$