You can only grow if you’re willing to feel awkward and uncomfortable when you try something new, Brian Tracy

An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:

**Dependent and independent variables**. Variables represent unknown quantities. The independent variable is chosen freely, while the dependent variable changes in response to the independent variable.**Constants**. Fixed numerical values that do not change.**Algebraic operations**. Operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction.

Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

It involves (e.g., $\frac{dy}{dx} = 3x +5y$):

**Dependent variables**:*Variables that depend on one or more other variables*(y).**Independent variables**: Variables upon which the dependent variables depend (x).**Derivatives**: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

- The function f(x, y) (the right-hand side of the ODE) in y’ = f(x, y) is continuous in a neighborhood around a point (x
_{0}, y_{0}) and - Its partial derivative with respect to y, $\frac{∂f}{∂y}$, is also continuous near (x
_{0}, y_{0}).

Then the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x_{0}, y_{0}) .

A first-order linear differential equation (ODE) has the general form: a(x)y' + b(x)y = c(x) where y′ is the derivative of y with respect to x, and a(x), b(x), and c(x) are functions of x. If c(x) = 0, the equation is called homogeneous, i.e., a(x)y’ + b(x)y = 0.

The equation can also be written in the standard linear form as: y’ + p(x)y = q(x) where $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$

In advanced calculus, complex functions play a significant role in solving differential equations and evaluating integrals, especially those involving trigonometric functions.

A **complex number** is *any number that can be expressed or written in the form z = a + bi*, where a and b are real numbers, and i is the imaginary unit, defined by the property i^{2} = −1.

In the complex number z = a + bi: a is called the real part, denoted as Re(z) = a. b is called the imaginary part, denoted as Im(z) = b.

**Example**: z = 3 +4i, the real part is Re(z) = 3 and the imaginary part is Im(z) = 4.

The complex conjugate of a complex number z = a + bi is denoted by $\bar z$ and is defined as $\bar z = a -bi$.

Multiplying a complex number by its conjugate results in a real number: $z·\bar z = (a + bi)(a -bi) = a^2 + b^2$, e.g., z = 3 + 4i, $z·\bar z = (3 + 4i)(3 - 4i) = 3^2 + 4^2 = 25.$

To divide complex numbers, eliminate the imaginary unit from the denominator by multiplying the numerator and denominator by the complex conjugate of the denominator.

Example: Calculate $\frac{2+i}{1-3i}$

$\frac{2+i}{1-3i}$ =[To simplify the expression, multiply both numerator and denominator by the conjugate of the denominator:] $\frac{2+i}{1-3i}·\frac{1+3i}{1+3i} = \frac{-1+7i}{10} =[\text{Thus, the simplified form is:}] \frac{-1}{10}+\frac{7}{10}i$.

A complex number z = a +bi can also be expressed in polar form, which uses its modulus r and argument θ:

- The modulus r = |z| = $\sqrt{a^2+b^2}$. It represents the distance of the complex number from the origin in the complex plane.
- The argument $θ = arg(z) = tan^{-1}(\frac{b}{a})$. It is the angle between the positive real axis and the line representing the complex number. Ensure the correct quadrant for θ by considering the signs of a and b.

The polar form of a complex number is: z = a + bi = rcos(θ) + rsin(θ)i = r(cos(θ) + isin(θ)) =[This can be written compactly using Euler’s formula] re^{iθ}

Euler’s formula is a fundamental formula in complex analysis. It establishes the mathematical relationship between trigonometric functions and the complex exponential function. It states that e^{iθ} = cos(θ) + isin(θ). This allows us to represent complex numbers compactly in exponential form.

The laws of exponents apply to complex numbers in exponential form, similar to real numbers.

- Product Rule: a
^{m}·a^{n}= a^{m+n}. - Quotient Rule: a
^{m}÷a^{n}= a^{m-n} - Power of a Power Rule: (a
^{m})^{n}= a^{mn} - Power of a Product Rule: (ab)
^{m}= a^{m}b^{m} - Power of a Quotient Rule: (a/b)
^{m}= (a^{m})/(b^{m}) - Zero Exponent Rule: a
^{0}= 1 for a ≠ 0 - Negative Exponent Rule: a
^{-m}= 1/(a^{m}) - Fractional Exponent Rule: $a^{\frac{m}{n}} = \sqrt[n]{a^m}$

Multiplying complex numbers in polar form is simplified using exponentials.

Given $z_1 = r_1e^{iθ_1},~ z_2 = r_2e^{iθ_2}$, $z_1·z_2 = r_1r_2e^{iθ_1}·e^{iθ_2}=r_1r_2e^{i(θ_1+θ_2)}$. This shows that:

- The modulus of the product is the product of the moduli: |z
_{1}z_{2}| = r_{1}r_{2}. - The argument of the product is the sum of the arguments, arg(z
_{1}z_{2}) = θ_{1}+ θ_{2}

Proof:

$z_1·z_2 = r_1r_2e^{iθ_1}·e^{iθ_2}=[\text{Laws of Exponents. Product Rule.}]r_1r_2e^{iθ_1+iθ_2}=r_1r_2e^{i(θ_1+θ_2)}$

**Differentiation of Complex Functions**
A complex-valued function can be expressed in terms of its real and imaginary parts. For a function u(t) + iv(t), where:

- u(t) is the real part, a real-valued function of t.
- v(t) is the imaginary part, a real-valued function of t.
- i is the imaginary unit, with i
^{2}= -1.

**Differentiation Rule**. The derivative of y(t) with respect to t is obtained by differentiating the real and imaginary parts separately: $\frac{d}{dt}y(t) = \frac{d}{dt}[u(t)+iv(t)] = \frac{du}{dt}+i\frac{dv}{dt}$ ↭ D(u + iv)= Du + iDv. This rule follows from the linearity of differentiation.

Example: Differentiating the Complex Exponential Function. Consider the function y(t) = e^{it}. $\frac{d}{dt}e^{it} =[\text{Using Euler’s formula}] \frac{d}{dt}(cos(t) + isin(t)) =[\text{Differentiation Rule}] \frac{d}{dt}(cos(t)) + i\frac{d}{dt}(sin(t)) = -sin(t) + icos(t) = i(cos(t) + isin(t)) =[\text{Using Euler’s formula}] ie^{it}$.

$\frac{d}{dt}e^{it} = ie^{it}$. The derivate of e^{it} is ie^{it} because multiplying by i rotates the complex number e^{it} by 90° in the complex plane. This property is fundamental in solving differential equations involving complex exponentials.

Consider the differential equation: $\frac{dy}{dt} = iy$ where y is a function of t, and i is the imaginary unit with the initial condition y(0) = 1.

**General Solution**.

This is a first-order linear ordinary differential equation (ODE). We can solve it using the method of separation of variables.

$\frac{dy}{dt} = iy ↭[\text{Separate variables}] \frac{dy}{y} = idt ↭[\text{Integrate both sides}] \int \frac{dy}{y} = \int idt ↭ lny = it + C$ where C is the constant of integration. Solve for y: $y = e^{it+C} = e^Ce^{it}$. Let K = e^{C}, the general solution is y(t) = Ke^{it}.

Applying the initial condition. Given y(0) = 1 ↭ y(0) = Ke^{i·0} = K·1 = K = 1. Final solution: y(t) = e^{it}

Verification: $\frac{d}{dt}y(t) = \frac{d}{dt}e^{it} = ie^{it} = i(y(t)), y(0)=e^{i·0} = cos(0) + isin(0) = 1 + i·0 = 1$. This confirms that y(t) = e^{it} satisfies the differential equation and the initial condition.

Understanding complex exponents is crucial for working with complex functions.

**General Expression**. Consider the general exponential expression e^{a+ib} where a and b are real numbers. e^{a+ib}.

e^{a+ib} =[Using properties of exponents, this expression can be rewritten as a product of two simpler exponentials] e^{a}·e^{ib} where e^{a} is a real number and e^{ib} is a complex exponential, representing rotation in the complex plane.

Euler’s formula relates complex exponentials to trigonometric functions: e^{ib} = cos(b) + isin(b).

Combining the expressions, the complex exponential e^{a+b} can be written as: e^{a+ib} = e^{a}(cos(b) + isin(b)).

e^{a} is the magnitude or modulus, and b is the angle. It represents the rotation angle in the complex plane.

Complex numbers can be represented in polar form, which is particularly useful for multiplication and division.

Given a complex number α = e^{a+ib}. It can also be represented in polar form as re^{iθ} = r(cos(θ) + isin(θ)) (Refer to Figure iii for a visual representation and aid in understanding it), where:

- r = |α| = e
^{a}is the modulus or magnitude of the complex number. It is the distance from the origin to the point α in the complex plane. - θ = arg(α) = b is the argument or the angle between the positive real axis and the line representing α.

Integrating complex functions can simplify calculations involving real integrals, especially when they include trigonometric functions.

Consider the integral: $\int e^{-x}cos(x)dx$. We can use complex exponentials to simplify the integral by expressing cos(x) as the real part of e^{ix}, cos(x) = $Re(e^{ix})$

$\int e^{-x}cos(x)dx$ =[Express the integral in terms of e^{ix}] $Real(\int e^{-x}e^{ix}dx) = Real(\int e^{(-1+i)x}dx)$

Integrate the complex exponential: $\int e^{(-1+i)x}dx = \frac{e^{(-1+i)x}}{-1+i} + C$ where C is the constant of integration.

Simplifying the denominator: $\frac{e^{(-1+i)x}}{-1+i} = \frac{1}{-1+i}e^{-x}e^{ix} = \frac{1}{-1+i}e^{-x}(cos(x)+isin(x))=\frac{1}{-1+i}\frac{-1-i}{-1-i}e^{-x}(cos(x)+isin(x)) = \frac{-1-i}{2}e^{-x}(cos(x)+isin(x))$ 🚀

$\int e^{-x}cos(x)dx = Real (\frac{e^{(-1+i)x}}{-1+i}) =[\text{Taking the real part 🚀}] \frac{e^{-x}}{2}(-cos(x)+sin(x))+C$

To find all n-th roots of unity, we solve the equation: x^{n} = 1 where z is a complex number. This can be approached using Euler’s formula, which connects complex exponentials with trigonometric functions.

In the real number system, the only solution to this equation is x = 1, because it is the only real number that, when raised to any power, results in 1.

However, in the complex domain, the number 1 has n different n-th roots which lie in the complex plane. These roots are distributed evenly on the unit circle (the set of points in the complex plane at a distance of 1 from the origin).

The number 1 can be represented using Euler’s formula by choosing θ = 0: 1 = $e^{i·0} = cos(θ)+ isin(θ) = 1 + i·0$. The complex equation z^{n} = 1 can be rewritten using exponential form: $z^n = e^{i·0} = e^{i·2πk}$ for any integer k because adding multiples of 2π to the angle 0 does not change the value of e^{iθ} due to the periodicity of sine and cosine functions.

$z^n = e^{i·2πk}$ ⇒[Solving for z:] $z = e^{\frac{i·2πk}{n}}$ for k = 0, 1, 2, ···, n -1. There are n distinct solutions corresponding to k = 0 through k = n-1. Using Euler’s formula, each root can be expressed as: $z_k = cos(\frac{i·2πk}{n})+isin(\frac{i·2πk}{n})$ where k = 0, 1, 2, ···, n -1.

Conclusion: The n-th roots of unity are the solution to the equation z^{n} = 1, where z is a complex number. These roots are given by the formula: $z_k = e^{\frac{2πik}{n}} = cos(\frac{2πk}{n}) + isin(\frac{2πk}{n})$ where k = 0, 1, 2, ···, n-1 and:

- $e^{\frac{2πik}{n}}$ represents the exponential form of complex numbers, derived from Euler’s formula e
^{iθ}= cos(θ) + isin(θ). - k is the index that determines the specific root.
- These roots are evenly spaced around the unit circle in the complex plane at angles that are multiples of $\frac{2π}{n}$

**Square Roots of Unity (n = 2)**: z^{2}= 1.

k = 0, $z_0 = e^{\frac{2πi·0}{2}}=e^0=1$. k = 1, $z_1 = e^{\frac{2πi·1}{2}}=e^{πi} = cos(π)+isin(π) = -1 + i·0 = -1.$

The square roots of unity are 1 and −1, lying on opposite ends of the real axis on the unit circle.

**Fourth root of Unity (n = 4)**: z^{4}= 1.

k = 0, $z_0 = e^{\frac{2πi·0}{4}} = e^0 = 1$. k = 1, $z_1 = e^{\frac{2πi·1}{4}} = e^{\frac{πi}{2}} = cos(\frac{π}{2})+isin(\frac{π}{2}) = 0 + i·1 = i$. k = 2, $z_2 = e^{\frac{2πi·2}{4}}=e^{πi} = cos(π)+isin(π) = -1 + i·0 = -1$. k = 3, $z_3 = e^{\frac{2πi·3}{4}} = e^{\frac{3πi}{2}} = cos(\frac{3π}{2})+isin(\frac{3π}{2}) = 0 + i·(-1) = -i$

The fourth roots of unity are 1, i, −1, and -i, which are positioned at equal angles of ^{π}⁄_{2} radians around the unit circle.

In the complex plane, the n-th roots of unity are arranged as the vertices of a regular n-sided polygon inscribed in the unit circle. Each vertex corresponds to one of the roots, and the angle between any two consecutive roots is $\frac{2π}{n}$ radians (Refer to Figure iv for a visual representation and aid in understanding it).

Properties of N-th Roots of Unity:

- All n roots z
_{0}, z_{1}, ··, z_{n-1}are distinct. - Each root is obtained by rotating the previous one by $\frac{2π}{n}$ radians.
- The product of any two n-th roots of unity is also an n-th root of unity: $z_k·z_m = e^{\frac{2πik}{n}}e^{\frac{2πim}{n}} = e^{\frac{2πi(k+m)}{n}} = z_{(k + m)~ mod~ n}$
- Each n-th root of unity has an inverse, which is its complex conjugate: $z_k^{-1} = \bar z_k = e^{-\frac{2πik}{n}} = z_{n-k}$.
- The sum of all n-th roots of unity is zero: $\sum_{k=0}^{n-1} z_k = 0$.
They form the vertices of a regular polygon centered at the origin, and their vector sum cancels out.

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].

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