1.3 Q-3 | Complex Numbers | Mathematics 11

Exercise 1.3 — Question 3

Factorize the following into linear factors:

(i) $3 z^{2} + 363$

(ii) $z^{2} + 5$

(iii) $9 z^{2} + 16$

Over the real numbers, $z^{2} + a^{2}$ cannot be factorized. However, over $C$ , using $i^{2} = - 1$ :

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

This is the difference of squares identity applied with complex numbers.

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

Apply the sum of squares factorization with $a = 5$ : $z^{2} + 5 = z^{2} - (5 i)^{2} = (z + 5 i) (z - 5 i)$

Result: $z^{2} + 5 = (z + 5 i) (z - 5 i)$

Step 1: Rewrite as a perfect square form: $9 z^{2} + 16 = (3 z)^{2} + 4^{2}$

Step 2: Apply the sum of squares factorization: $(3 z)^{2} + 4^{2} = (3 z)^{2} - (4 i)^{2} = (3 z + 4 i) (3 z - 4 i)$

Result: $9 z^{2} + 16 = (3 z + 4 i) (3 z - 4 i)$

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