Precalculus by Richard Wright
Jesus answered, “I am the way and the truth and the life. No one comes to the Father except through me.” John 14:6 NIV
Summary: In this section, you will:
SDA NAD Content Standards (2018): PC.5.6
Suppose you need to describe locations to completely and evenly encircle something. Let’s say you have exactly five people and you want to have them spaced evenly around a circle with radius of 2. You could describe these points in the complex plane as \(\sqrt[5]{32}\). We will look at this later in the lesson, but first let’s look at other operations with complex numbers.
In standard form (a + bi), multiplying complex numbers involved using the distributive property and changing i2 into −1.
(2 + i)(3 − 2i)
6 − 4i + 3i − 2i2
6 − 4i + 3i + 2
8 − i
However, having the complex numbers in trigonometric form (r(cos θ + i sin θ)) makes the multiplication much easier. Simply multiply the r’s and add the angles.
Let z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ2 + i sin θ2) be complex numbers.
z1z2 = r1r2(cos(θ1 + θ2) + i sin(θ1 + θ2))
If θ1 + θ2 > 2π, then subtract 2π to get a coterminal angle between 0 and 2π.
If \(z_1 = 3\left(\cos \frac{π}{3} + i \sin \frac{π}{3}\right)\) and \(z_2 = 12\left(\cos \frac{π}{6} + i \sin \frac{π}{6}\right)\), find z1z2.
Solution
$$z_1 z_2 = r_1 r_2\left(\cos \left(θ_1 + θ_2\right) + i \sin \left(θ_1 + θ_2\right)\right)$$
$$= 3 \cdot 12\left(\cos \left(\frac{π}{3} + \frac{π}{6}\right) + i \left(\frac{π}{3} + \frac{π}{6}\right)\right)$$
$$= 36\left(\cos \frac{π}{2} + i \sin \frac{π}{2}\right)$$
If \(z_1 = 2\left(\cos \frac{5π}{6} + i \sin \frac{5π}{6}\right)\) and \(z_2 = 5\left(\cos \frac{π}{2} + i \sin \frac{π}{2}\right)\), find z1z2.
Answer
\(10\left(\cos \frac{4π}{3} + i \sin \frac{4π}{3}\right)\)
In standard form, division involved multiplying the numerator and denominator by the complex conjugate of the denominator.
$$\frac{a + bi}{c + di} = \frac{\left(a + bi\right)\left(c - di\right)}{\left(c + di\right)\left(c - di\right)}$$
In trigonometric form, simply divide the r’s and subtract the angles.
Let \(z_1 = r_1\left(\cos θ_1 + i \sin θ_1\right)\) and \(z_2 = r_2\left(\cos θ_2 + i \sin θ_2\right)\) be complex numbers.
$$\frac{z_1}{z_2} = \frac{r_1}{r_2}\left(\cos \left(θ_1 - θ_2\right) + i \sin \left(θ_1 - θ_2\right)\right)$$
If θ1 − θ2 < 0, then add 2π to get a coterminal angle between 0 and 2π.
If \(z_1 = 3\left(\cos \frac{π}{3} + i \sin \frac{π}{3}\right)\) and \(z_2 = 12\left(\cos \frac{π}{6} + i \sin \frac{π}{6}\right)\), find \(\frac{z_1}{z_2}\).
Solution
$$\frac{z_1}{z_2} = \frac{r_1}{r_2}\left(\cos \left(θ_1 - θ_2\right) + i \sin \left(θ_1 - θ_2\right)\right)$$
$$= \frac{3}{12}\left(\cos \left(\frac{π}{3} - \frac{π}{6}\right) + i \left(\frac{π}{3} - \frac{π}{6}\right)\right)$$
$$= \frac{1}{4}\left(\cos \frac{π}{6} + i \sin \frac{π}{6}\right)$$
If \(z_1 = 2\left(\cos \frac{5π}{6} + i \sin \frac{5π}{6}\right)\) and \(z_2 = 5\left(\cos \frac{π}{2} + i \sin \frac{π}{2}\right)\), find \(\frac{z_1}{z_2}\).
Answer
\(\frac{2}{5}\left(\cos \frac{π}{3} + i \sin \frac{π}{3}\right)\)
The only way to deal with exponents with complex numbers in standard form is repeated multiplication which can be tedious. Exponents in trigonometric form are much simpler. Since the multiplication process is multiplying the r’s and adding the angles, exponents which are repeated multiplication will repeatedly multiply the r’s (or exponent) and repeated add the angles (or multiply).
Let z = r(cos θ + i sin θ) be a complex number.
zn = rn(cos(nθ) + i sin(nθ))
If nθ > 2π, then subtract 2π to get a coterminal angle between 0 and 2π.
If \(z = 3\left(\cos \frac{π}{3} + i \sin \frac{π}{3}\right)\), find z4.
Solution
$$z^n = r^n\left(\cos \left(nθ\right) + i \sin \left(nθ\right)\right)$$
$$= 3^4\left(\cos \left(4 \frac{π}{3}\right) + i \sin \left(4 \frac{π}{3}\right)\right)$$
$$= 81\left(\cos \frac{4π}{3} + i \sin \frac{4π}{3}\right)$$
If \(z = 2\left(\cos \frac{π}{6} + i \sin \frac{π}{6}\right)\), find z3.
Answer
\(8\left(\cos \frac{π}{2} + i \sin \frac{π}{2}\right)\)
Roots with complex numbers in standard form has not been previously addressed in this book; however, trigonometric form makes it manageable.
First consider a simple example. You would probably solve this equation this way.
x4 − 16 = 0
x4 = 16
$$x = ± \sqrt[4]{16}$$
x = −2, 2
However, the fundamental theorem of algebra says that there should be 4 solutions not 2. Try solving by factoring.
x4 − 16 = 0
(x2 − 4)(x2 + 4) = 0
(x − 2)(x + 2)(x2 + 4) = 0
x – 2 = 0 | x + 2 = 0 | x2 + 4 = 0 |
x = 2 | x = −2 | x2 = -4 |
\(x = ±\sqrt{-4}\) | ||
x = −2i, 2i |
This gives all four solutions. Notice that if you graph these four solutions on a complex plane, they are evenly spaced around a circle with radius \(\sqrt[4]{16} = 2\).
To find these roots for any complex numbers, find the radius of the circle and then the evenly spaced angles around the circle. Exponents had rn and nθ. Roots are the inverse of exponents so the process should be also be the inverses for r and θ.
$$\sqrt[n]{r}\left(\cos \frac{θ}{n} + i \sin \frac{θ}{n}\right)$$
gives the first root.
To get the rest of the roots, divide the circle into n parts and add that to the angle.
Let \(z = r\left(\cos θ + i \sin θ\right)\)
$$\sqrt[n]{z} = \sqrt[n]{r}\left(\cos \left(\frac{θ}{n} + \frac{2πk}{n}\right) + i \sin \left(\frac{θ}{n} + \frac{2πk}{n}\right)\right)$$
Where k = 0, 1, 2, …, n – 1
Find the cube roots of z = 4 – 4i. Write the answers in standard form.
Solution
First, write the complex number in trigonometric form. Find r.
$$r = \sqrt{a^2 + b^2}$$
$$r = \sqrt{4^2 + \left(-4\right)^2}$$
$$r = 4\sqrt{2}$$
Find θ.
$$\tan θ = \frac{b}{a}$$
$$\tan θ = \frac{-4}{4} = -1$$
$$θ = \frac{7π}{4}$$
$$z = 4\sqrt{2} \left(\cos \frac{7π}{4} + i \sin \frac{7π}{4}\right)$$
Now, find the cube roots.
$$\sqrt[n]{z} = \sqrt[n]{r}\left(\cos \left(\frac{θ}{n} + \frac{2πk}{n}\right) + i \sin \left(\frac{θ}{n} + \frac{2πk}{n}\right)\right)$$
$$\sqrt[3]{z} = \sqrt[3]{4\sqrt{2}}\left(\cos \left(\frac{\frac{7π}{4}}{3} + \frac{2πk}{3}\right) + i \sin \left(\frac{\frac{7π}{4}}{3} + \frac{2πk}{3}\right)\right)$$
$$= \sqrt[3]{4\sqrt{2}}\left(\cos \left(\frac{7π}{12} + \frac{8πk}{12}\right) + i \sin \left(\frac{7π}{12} + \frac{8πk}{12}\right)\right)$$
Let k = 0:
$$\sqrt[3]{4\sqrt{2}}\left(\cos \frac{7π}{12} + i \sin \frac{7π}{12}\right)$$
$$≈ -0.4612 + 1.7211i$$
Let k = 1:
$$\sqrt[3]{4\sqrt{2}}\left(\cos \frac{15π}{12} + i \sin \frac{15π}{12}\right)$$
$$≈ -1.2599 - 1.2599i$$
Let k = 2:
$$\sqrt[3]{4\sqrt{2}}\left(\cos \frac{23π}{12} + i \sin \frac{23π}{12}\right)$$
$$≈ 1.7211 - 0.4612i$$
If \(z = 8\left(\cos \frac{π}{6} + i \sin \frac{π}{6}\right)\), find \(\sqrt[3]{z}\). Write your answers in standard form.
Answers
1.970 + 0.347i, −1.286 + 1.532i, −0.684 − 1.879i
Let z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ2 + i sin θ2) be complex numbers.
z1z2 = r1r2(cos(θ1 + θ2) + i sin(θ1 + θ2))
If θ1 + θ2 > 2π, then subtract 2π to get a coterminal angle between 0 and 2π.
Let z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ2 + i sin θ2) be complex numbers.
$$\frac{z_1}{z_2} = \frac{r_1}{r_2}\left(\cos \left(θ_1 - θ_2\right) + i \sin \left(θ_1 - θ_2\right)\right)$$
If θ1 − θ2 < 0, then add 2π to get a coterminal angle between 0 and 2π.
Let z = r(cos θ + i sin θ) be a complex number.
zn = rn(cos(nθ) + i sin(nθ))
If nθ > 2π, then subtract 2π to get a coterminal angle between 0 and 2π.
Let z = r(cos θ + i sin θ)
$$\sqrt[n]{z} = \sqrt[n]{r}\left(\cos \left(\frac{θ}{n} + \frac{2πk}{n}\right) + i \sin \left(\frac{θ}{n} + \frac{2πk}{n}\right)\right)$$
Where k = 0, 1, 2, …, n – 1
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