Algebra 2 by Richard Wright
Are you not my student and
has this helped you?
Objectives:
SDA NAD Content Standards (2018): AII.4.1, AII.4.2
Sometimes, solving a quadratic equation will give no real solution, but mathematicians decided to explore those solutions more. In the process, they gave them confusing names. Solutions that are on the number line are called real numbers. The solutions to quadratic equations that do not fall on the number line always included square roots of negative numbers. These are called imaginary numbers. Imaginary numbers can be used to generate art like the Mandelbrot set as in figure 1.
Imaginary numbers are defined new numbers are follows.
$$ \sqrt{-1}=i $$
$$ i^2=-1 $$
i is called the imaginary unit, and numbers with i are called imaginary numbers. That name does not mean these numbers are made up and worthless imaginations, but they are actually useful. Complex numbers are all the numbers that include real and imaginary numbers. They are written in the form
a + bi
where a is the real part and bi is the imaginary part.
Evaluate (a) \(\sqrt{-16}\) and (b) \(\sqrt{-24}\).
Solution
Split the square root into factors with −1 being one of the factors.
$$ \sqrt{-16}=\sqrt{16}\sqrt{-1} $$
Remember that \(\sqrt{-1}=i\).
= 4i
Split the square root into factors of a perfect square and −1 being factors.
$$ \sqrt{-24} = \sqrt{4} \sqrt{6} \sqrt{-1} $$
Remember that \(\sqrt{-1}=i\)
$$ = \mathbf{2\sqrt{6}i} $$
Complex numbers can be added, subtracted, multiplied, and divided.
To add and subtract complex numbers, combine like terms.
Simplify (a) (2 + 3i) + (1 – 4i) and (b) (5 – 7i) – (2i).
Solution
Combine the constant terms and the terms with i.
(2 + 3i) + (1 – 4i)
(2 + 1) + (3i + (–4i))
3 + (–i)
3 – i
Combine the constant terms and the terms with i. Remember to subtract this time.
(5 – 7i) – (2i)
(5) + (–7i – 2i)
5 – 9i
To multiply complex numbers, use the distributive property. Whenever an i2 is generated, change it to −1.
Simplify (a) (2 + 3i)(1 – 4i) and (b) (5 – 7i)(2i).
Solution
Distribute each term of the first complex number to both terms of the second complex number.
(2 + 3i)(1 – 4i)
2·1 + 2·–4i + 3i·1 + 3i·–4i
2 – 8i + 3i – 12i2
Change the i2 to –1.
2 – 8i + 3i – 12(–1)
2 – 8i + 3i + 12
14 – 5i
Distribute the 2i to both terms of the first complex number.
(5 – 7i)(2i)
5·2i – 7i·2i
10i – 14i2
Change the i2 to –1.
10i – 14(–1)
10i + 14
14 + 10i
The complex conjugate is the same complex number except for the opposite sign on the imaginary part. 2 + 3i and 2 – 3i are complex conjugates.
To divide complex number,
Multiply the numerator and denominator by the complex conjugate of the denominator.
This is multiplying by 1 and does not change the result except for how it looks; however, this does remove all the i's from the denominator.
In a simplified complex number, there are no i's in the denominator.
Simplify (a) \(\frac{2+3i}{1-4i}\) and (b) \(\frac{5-7i}{2i}\)
Solution
Multiply the numerator and denominator by the complex conjugate of the denominator.
$$ \frac{2+3i}{1-4i} $$
$$ \frac{\left(2+3i\right)\left(1+4i\right)}{\left(1-4i\right)\left(1+4i\right)} $$
Multiply.
$$ \frac{2+8i+3i+12i^2}{1+4i-4i-16i^2} $$
Replace the i2 with –1.
$$ \frac{2+8i+3i+12\left(-1\right)}{1+4i-4i-16\left(-1\right)} $$
Simplify the numerator and denominator.
$$ \frac{2+11i-12}{1+16} $$
$$ \frac{-10+11i}{17} $$
Put into standard form by separating into two fractions.
$$ \mathbf{-\frac{10}{17}+\frac{11}{17}i} $$
Multiply the numerator and denominator by the complex conjugate of the denominator.
$$ \frac{5-7i}{2i} $$
$$ \frac{\left(5-7i\right)\left(-2i\right)}{\left(2i\right)\left(-2i\right)} $$
Distribute.
$$ \frac{-10i+14i^2}{-4i^2} $$
Replace the i2 with –1.
$$ \frac{-10i+14\left(-1\right)}{-4\left(-1\right)} $$
Simply the numerator and denominator.
$$ \frac{-10i-14}{4} $$
$$ \frac{-14-10i}{4} $$
Put into standard form by separating into two fractions.
$$ -\frac{14}{4}-\frac{10}{4}i $$
Reduce.
$$ \mathbf{-\frac{7}{2}-\frac{5}{2}i} $$
105 #1, 3, 5, 7, 17, 19, 21, 23, 25, 29, 31, 33, 35, 37, 39, 43, 49, 51 and division and mixed review = 25
Divide.
Mixed Review
; (2, 0), (3, 0)