1.3 Exercise 1.3 — Factorization and Completing the Square

This exercise covers three key skills for working with polynomials and equations over $\mathbb{C}$:

1. Factorizing polynomials into linear factors over $\mathbb{C}$
2. Solving quadratic equations by completing the square
3. Solving simultaneous linear equations with complex coefficients

---

1. Factorization into Linear Factors over $\mathbb{C}$

Every polynomial can be factorized into linear factors over $\mathbb{C}$ (Fundamental Theorem of Algebra).

Sum of Squares

Using $i^2=-1$, a sum of squares becomes a difference of squares:

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

Example: Factorize $3z^2+363$

$3z^2+363=3(z^2+121)=3(z^2-(11i{)}^2)=3(z+11i)(z-11i)$

Cubic Polynomials

To factorize a cubic $P(z)=z^3+b{z}^2+cz+d$:

1. Use the Factor Theorem: test divisors of the constant term $d$ to find $z=k$ such that $P(k)=0$. Then $(z-k)$ is a factor.
2. Perform synthetic division to reduce to a quadratic.
3. Factorize the quadratic (using the quadratic formula if needed).

Example: Factorize $z^3-7z+6$

• Test $z=1$: $1-7+6=0$ ✓ → $(z-1)$ is a factor
• Synthetic division gives: $z^3-7z+6=(z-1)(z^2+z-6)$
• Factorize quadratic: $z^2+z-6=(z+3)(z-2)$
• Result: $(z-1)(z+3)(z-2)$

---

2. Solving Quadratic Equations by Completing the Square

For $p{z}^2+qz+r=0$ where $p,q,r\in \mathbb{R}$:

Method:

1. Divide through by $p$ (if $p\ne 1$)
2. Move the constant to the right side
3. Add ${\left(\frac{q}{2p}\right)}^2$ to both sides
4. Write the left side as a perfect square
5. Take square roots (both $\pm$), allowing complex values

Example: Solve $z^2-6z+2=0$

$z^2-6z=-2$ $z^2-6z+9=-2+9$ $(z-3{)}^2=7$ $z-3=\pm \sqrt{7}$ $z=3\pm \sqrt{7}$

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

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

---

3. Solving Simultaneous Linear Equations with Complex Coefficients

The standard methods (substitution, elimination) apply, but we must handle complex arithmetic carefully.

Key Technique: Removing $i$ from Denominators

If a term has $i$ in the denominator, multiply numerator and denominator by $i$:

$\frac{3}{i}=\frac{3\cdot i}{i\cdot i}=\frac{3i}{-1}=-3i$

More generally, to divide by a complex number $a+bi$, multiply by its conjugate $a-bi$:

$\frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}$

Example: Solve the system

$z+iw=2(1)$ $iz-w=3(2)$

Step 1: From equation (1): $z=2-iw$

Step 2: Substitute into (2): $i(2-iw)-w=3$ $2i-i^{2}w-w=3$ $2i+w-w=3\text{(since }{i}^2=-1\text{)}$

Wait — let's redo carefully: $2i-i^{2}w-w=3\Rightarrow 2i+w-w=3$

Actually: $-i^{2}w-w=w-w=0$... Let's use elimination instead.

Elimination method:

Multiply (1) by $i$: $iz+i^{2}w=2i\Rightarrow iz-w=2i$

Subtract from (2): $(iz-w)-(iz-w)=3-2i$...

Multiply equation (1) by $i$: $iz-w=2i(1^{\prime })$

Equation (2) is: $iz-w=3$

Subtracting (1') from (2): $0=3-2i$ — contradiction, so let's use the original correctly.

Correct approach — substitution:

From (2): $w=iz-3$

Substitute into (1): $z+i(iz-3)=2$ $z+i^{2}z-3i=2$ $z-z-3i=2$ $-3i=2$

This system is inconsistent for these specific values. For a consistent example:

Example: Solve $(1-i)z+(1+i)w=3$ and $(1+i)z+(1-i)w=1+2i$

Add the two equations: $[(1-i)+(1+i)]z+[(1+i)+(1-i)]w=4+2i$ $2z+2w=4+2i$ $z+w=2+i(\ast )$

Subtract equation 2 from equation 1: $[(1-i)-(1+i)]z+[(1+i)-(1-i)]w=2-2i$ $-2iz+2iw=2-2i$ $-z+w=\frac{2-2i}{2i}=\frac{(2-2i)(-i)}{2}=\frac{-2i+2i^2}{2}=\frac{-2-2i}{2}=-1-i$ $-z+w=-1-i(\ast \ast )$

From (*) and (**): Adding: $2w=1$, so $w=\frac{1}{2}$; then $z=2+i-\frac{1}{2}=\frac{3}{2}+i$.