1.3 Q-4

Exercise 1.3 — Question 4

This question covers factorization of polynomials over $\mathbb{C}$ (SLO M-11-A-08) and solving quadratic equations by completing the square (SLO M-11-A-09).

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Key Concept: Factorizing Sum of Squares

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

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

This is because $(ai{)}^2=a^2{i}^2=-a^2$, so $-(ai{)}^2=a^2$.

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Worked Examples: Factorization into Linear Factors

Example 1: Factorize $z^2+4$

$z^2+4=z^2-(2i{)}^2=(z+2i)(z-2i)$

Example 2: Factorize $z^2+9$

$z^2+9=z^2-(3i{)}^2=(z+3i)(z-3i)$

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

Step 1 — Factor out the GCD: $3z^2+363=3(z^2+121)$

Step 2 — Apply sum-of-squares factorization: $=3(z^2-(11i{)}^2)=3(z+11i)(z-11i)$

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Key Concept: Completing the Square

To solve $z^2+bz+c=0$ by completing the square:

1. Move the constant to the right: $z^2+bz=-c$
2. Add ${\left(\frac{b}{2}\right)}^2$ to both sides: $z^2+bz+{\left(\frac{b}{2}\right)}^2={\left(\frac{b}{2}\right)}^2-c$
3. Write the left side as a perfect square: ${\left(z+\frac{b}{2}\right)}^2={\left(\frac{b}{2}\right)}^2-c$
4. Take square roots (using $\sqrt{-1}=i$ if the right side is negative).

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Worked Examples: Completing the Square

Example 4: Solve $z^2+6z+13=0$

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

Example 5: Solve $z^2+4z+13=0$

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