LESSON 1

Solve the following equations

(a) $$x^2+1=0$$
(b) $$x^2+3x+3=0$$
(c) $$4x^2-2x=-1$$
SOLUTION
We need to use the quadratic formula
$$\displaystyle x={-b±\sqrt{b^2-4ac}\over 2a}$$
to solve these equations.
(a) $$x^2+1=0$$
$$x^2=-1$$
$$x=±\sqrt{-1}$$
$$x=±i$$

(b) $$x^2+3x+3=0$$
$$\displaystyle x={-3±\sqrt{3^2-4(1)(3)}\over 2(1)}$$
$$\displaystyle x={-3±\sqrt{-3}\over 2}$$
$$\displaystyle x={-3±\sqrt{3} i\over 2}$$

(c) $$4x^2-2x=-1$$
$$4x^2-2x+1=0$$
$$\displaystyle x={-(-2)±\sqrt{(-2)^2-4(4)(1)}\over 2(4)}$$
$$\displaystyle x={2±\sqrt{-12}\over8}$$
$$\displaystyle x={2±\sqrt{4(3)(-1)}\over8}$$
$$\displaystyle x={2±2\sqrt{3}i\over 8}$$
$$\displaystyle x={1\over4}±{\sqrt{3}\over4}i$$

EQUATIONS WITH COMPLEX COEFFICIENTS
LESSON 2
Determine $$z$$ such that
$$z^2+(2+2i)z-(15-10i)=0$$
SOLUTION
Again we will use the quadratic formula to solve the given equation.
$$z^2+(2+2i)z-(15-10i)=0$$
$$z^2+(2+2i)z-15+10i=0$$
$$\displaystyle z={-(2+2i)±\sqrt{(2+2i)^2-4(1)(-15+10i)}\over2(1)}$$
$$\displaystyle z={-(2+2i)±\sqrt{(2+2i)(2+2i)+60-40i}\over 2}$$
$$\displaystyle z={-(2+2i)±\sqrt{8i+60-40i}\over2}$$
$$\displaystyle z={-(2+2i)±\sqrt{60-32i}\over2}$$
• At this stage we need to determine the square root of $$60-32i$$ using the usual process.
$$\sqrt{60-32i}=a+bi$$
$$60-32i=(a+bi)^2$$
$$60-32i=a^2-b^2+2abi$$
$$a^2-b^2=60$$
$$\displaystyle ab=-16 ⟹ a=-{16\over b}$$

$$\displaystyle \left(-{16\over b}\right)^2-b^2=60$$
$$\displaystyle {256\over b^2} -b^2=60$$
$$b^4+60b^2-256=0$$
$$(b^2+64)(b^2-4)=0$$
$$b^2=4$$
$$b=±2$$

When $$b=2$$
$$\displaystyle a=-{16\over2}=-8$$
$$-8+2i$$

When $$b=-2$$
$$\displaystyle a=-{16\over -2}=8$$
$$8-2i$$
• We only need to substitute ONE of the complex square roots
$$\displaystyle z={-(2+2i)±(-8+2i)\over 2}$$
$$\displaystyle z={-(2+2i)+(-8+2i)\over 2}=-5$$
$$\displaystyle z={-(2+2i)-(-8+2i)\over2}=3-2i$$