M203 20260228 Trig and Complex Numbers

1. Polar form of Complex Numbers

1 ) Write the following in Polar Form:

Answers

2 ) Let $w = 3 + \sqrt{3}i$ and $z = 2\sqrt{3} - 6i$

• Find the relationship between $w$, $z$ and $wz$ in polar form

• Find the relationship between $w$, $z$ and $\dfrac{w}{z}$ in polar form

• Let $w = R(\cos\alpha + i\sin\alpha)$, $z = S(\cos\beta + i\sin\beta)$, what is $wz$? What is $\dfrac{w}{z}$?

Answer:

$wz = RS[\cos(\alpha + \beta) + i\sin(\alpha + \beta)]$

$\dfrac{w}{z} = \dfrac{R}{S}[\cos(\alpha - \beta) + i\sin(\alpha - \beta)]$

3 ) Prove De Moivre's Theorem:

#De_Moivres_Theorem

• $\LARGE (\cos \alpha + i \sin \alpha)^n = \cos(n\alpha) + i \sin(n\alpha)$

• $\LARGE (\cos \alpha - i \sin \alpha)^n = \cos(n\alpha) - i \sin(n\alpha)$

2. Exponential Form of Complex Numbers

Euler's Theorem

#Eulers_Theorem

$\LARGE \cos\alpha + i\sin\alpha = \LARGE e^{i\alpha}$

Express in Exponential Form or Rectangular Form:

• $-\dfrac{3}{\sqrt{3}} + 3i$

• $e^{\pi i}$

#Exponential_Form #Polar_Form

FormRepresentationComponents
Rectangular Form$x + yi$Rectangular coordinates $(x, y)$
Polar Form$R(\cos\theta + i\sin\theta)$ or $R \text{ cis } \theta$Polar coordinates $(R, \theta)$
Exponential Form$R e^{i\theta}$Magnitude $R$ and phase $\theta$

Express in Exponential Form or Rectangular Form:

• $\LARGE 6e^{\frac{\pi i}{3}}$

• $(1+i)^{20}$

For what value of $\alpha$ is $e^{i\alpha}=1$?

If $z = r e^{i \alpha}$, what is the conjugate of $z$?

Product of $z$ and its Conjugate

Multiplying a complex number by its conjugate results in the square of its magnitude (or modulus):

$$z \cdot \bar{z} = (r e^{i \alpha})(r e^{-i \alpha}) = r^2 e^0 = r^2$$

In rectangular form, if $z = x + yi$, this is equivalent to:

$$z \cdot \bar{z} = x^2 + y^2$$

Find $\cos\alpha$ and $\sin\alpha$ with $e^{i\alpha}$ and $e^{-i\alpha}$?

Euler's Inverse Formulas

#Eulers_Inverse_Formula

$$\cos\alpha = \frac{e^{i\alpha} + e^{-i\alpha}}{2}$$

$$\sin\alpha = \frac{e^{i\alpha} - e^{-i\alpha}}{2i}$$

an application of Euler's Inverse Formula for cosine to simplify an infinite series.

Mathematical Breakdown

The expression shows how to convert a trigonometric series into a geometric series using complex exponentials:

1. Euler’s Substitution: Using the identity $\cos(n\theta) = \frac{e^{in\theta} + e^{-in\theta}}{2}$, the term $\frac{\cos(n\theta)}{2^n}$ is rewritten.

2. Series Expansion: The summation is split into two separate parts:

$$\sum_{n=1}^{\infty} \frac{\cos(n\theta)}{2^n} = \sum_{n=1}^{\infty} \left( \frac{e^{in\theta}}{2^{n+1}} + \frac{e^{-in\theta}}{2^{n+1}} \right)$$

3. Purpose: This transformation is a common technique in competitive math (like the AMC) and calculus to evaluate infinite sums. By converting the trigonometric terms into complex exponentials, the expression becomes a sum of two geometric series, which are much easier to solve using the formula $S = \frac{a}{1-r}$.

To complete the evaluation of the series $\sum_{n=1}^{\infty} \frac{\cos(n\theta)}{2^n}$, we can sum the two resulting geometric series identified in the previous step.

Step-by-Step Evaluation

1. Identify the Geometric Series:

The expression is split into two series with the general form $\sum_{n=1}^{\infty} \frac{1}{2} \left( \frac{e^{i\theta}}{2} \right)^n$ and $\sum_{n=1}^{\infty} \frac{1}{2} \left( \frac{e^{-i\theta}}{2} \right)^n$.

2. Apply the Sum Formula:

For a geometric series $\sum_{n=1}^{\infty} ar^n$, the sum is $S = \frac{ar}{1-r}$. Here, $a = \frac{1}{2}$ and the common ratios are $r_1 = \frac{e^{i\theta}}{2}$ and $r_2 = \frac{e^{-i\theta}}{2}$.

$$S = \frac{1}{2} \left( \frac{\frac{e^{i\theta}}{2}}{1 - \frac{e^{i\theta}}{2}} + \frac{\frac{e^{-i\theta}}{2}}{1 - \frac{e^{-i\theta}}{2}} \right) = \frac{1}{2} \left( \frac{e^{i\theta}}{2 - e^{i\theta}} + \frac{e^{-i\theta}}{2 - e^{-i\theta}} \right)$$

3. Find a Common Denominator:

Combining the fractions:

$$\frac{e^{i\theta}(2 - e^{-i\theta}) + e^{-i\theta}(2 - e^{i\theta})}{(2 - e^{i\theta})(2 - e^{-i\theta})}$$

4. Simplify Using Euler's Formula:

5. Final Result:

Multiply by the $\frac{1}{2}$ from the start:

$$\text{Sum} = \frac{2\cos\theta - 1}{5 - 4\cos\theta}$$

3. Roots of Unity

#Roots_Of_Unity

Derivation of the Roots of $F(x) = x^n - 1$

To find the roots of $F(x) = x^n - 1$, we solve the equation $x^n = 1$, where the solutions are known as the $n^{th}$ roots of unity.

1. Exponential Representation of 1

Using Euler's Formula, the number $1$ can be expressed in the complex plane as:

$$1 = e^{i(2\pi m)}$$

where $m$ is any integer.

2. Set up the Equation

Substitute the complex form into the root equation:

$$x^n = e^{i(2\pi m)}$$

3. Solve for $x$

Take the $n^{th}$ root of both sides by raising them to the power of $1/n$:

$$x = (e^{i(2\pi m)})^{1/n}$$

$$x = e^{i\left(\frac{2\pi m}{n}\right)}$$

4. Define the Unique Roots

To obtain $n$ distinct roots, we let $m$ range from $0$ to $n-1$:

$$x_m = e^{i\left(\frac{2\pi m}{n}\right)} \quad \text{for } m = 0, 1, 2, \dots, n-1$$

5. Conversion to Polar Form

Using the identity $e^{i\theta} = \cos\theta + i\sin\theta$, the roots are:

$$x_m = \cos\left(\frac{2\pi m}{n}\right) + i\sin\left(\frac{2\pi m}{n}\right)$$

Here are the solutions for the roots of the polynomials presented in the provided images:

Solutions for Roots of $F(x) = x^n - 1$

PolynomialFactored FormRoots ($x$)Visualization
$F(x) = x^3 - 1$$(x-1)(x^2+x+1)$$1, -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i$$3^{\text{rd}}$ roots of unity
$F(x) = x^4 - 1$$(x^2-1)(x^2+1)$$1, -1, i, -i$$4^{\text{th}}$ roots of unity
$F(x) = x^6 - 1$$(x^3-1)(x^3+1)$$1, -1, \frac{1}{2} \pm \frac{\sqrt{3}}{2}i, -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i$$6^{\text{th}}$ roots of unity

Key Formulas Used

Would you like me to help Elaine's son work through the more advanced problem from the worksheet, such as solving for $x^6 = -64$?

Problem: Suppose $w$ is a root of $F(z) = z^n - 1$, what of the following must also be a root of $F(z)$: $\bar{w}$, $w-1$, $1/w$?

Explanation

This problem explores the properties of the $n^{\text{th}}$ roots of unity, which are the solutions to the equation $z^n = 1$.

1. The Conjugate ($\bar{w}$)

2. The Translation ($w-1$)

3. The Reciprocal ($1/w$)

Final Answer: Both $\bar{w}$ and $1/w$ must also be roots of $F(z)$.

2 ) If the six solutions of $x^6 = -64$ are written in the form $a+bi$, where $a$ and $b$ are real, then find the product of the solutions with $a>0$?

1. Represent $-64$ in Polar Form

In the complex plane, $-64$ lies on the negative real axis.

2. Solve for $x$

Take the $6^{\text{th}}$ root of both sides:

$$x = (64 e^{i(\pi + 2\pi m)})^{1/6}$$

$$x = 64^{1/6} \cdot e^{i\left(\frac{\pi + 2\pi m}{6}\right)}$$

Since $2^6 = 64$, the magnitude of each root is $2$:

$$x_m = 2 e^{i\left(\frac{\pi}{6} + \frac{\pi m}{3}\right)} \quad \text{for } m = 0, 1, 2, 3, 4, 5$$

3. List the Six Solutions

By plugging in $m$ and converting to $a + bi$ form ($x = 2(\cos \theta + i\sin \theta)$):

mAngle (θ)Solution (a+bi)
0$30^\circ$$2(\frac{\sqrt{3}}{2} + \frac{1}{2}i) = \sqrt{3} + i$
1$90^\circ$$2(0 + i) = 2i$
2$150^\circ$$2(-\frac{\sqrt{3}}{2} + \frac{1}{2}i) = -\sqrt{3} + i$
3$210^\circ$$2(-\frac{\sqrt{3}}{2} - \frac{1}{2}i) = -\sqrt{3} - i$
4$270^\circ$$2(0 - i) = -2i$
5$330^\circ$$2(\frac{\sqrt{3}}{2} - \frac{1}{2}i) = \sqrt{3} - i$

4. Find the Product of Solutions with $a > 0$

The question specifically asks for the product of solutions where the real part $a$ is positive. These are the solutions for $m=0$ and $m=5$:

$$(\sqrt{3} + i)(\sqrt{3} - i) = (\sqrt{3})^2 - (i)^2 = 3 - (-1) = 4$$

Find the roots of the polynomials

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4 ） Find all solutions to $z^6 + z^4 + z^3 + z^2 + 1 = 0$.

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