Complex Numbers : Generic Form of Complex Numbers

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Generic Form of Complex Numbers

what you'll learn...

Overview

Generic Form of Complex Numbers

» All complex numbers are in the form $a + i b$ $a+ib$

» Example: Solutions of $x^{3} = - 1$ $x^3=-1$

→ ${(\sqrt[3]{- 1})}_{1 s t} = e^{\frac{i π}{3}}$ $(root(3)(-1))_(1st) =e^((i pi)/3)$ $= \frac{1}{2} + i \frac{\sqrt{3}}{2}$ $= 1/2 + i sqrt(3)/2$

→ ${(\sqrt[3]{- 1})}_{2 n d} = e^{i π}$ $(root(3)(-1))_(2nd) =e^(i pi)$ $= - 1$ $= -1$

→ ${(\sqrt[3]{- 1})}_{3 r d} = e^{\frac{i 5 π}{3}}$ $(root(3)(-1))_(3rd) =e^((i 5pi)/3)$ $= \frac{1}{2} - i \frac{\sqrt{3}}{2}$ $= 1/2 - i sqrt(3)/2$

recap quadratic equation

A quadratic equation of the form $p x^{2} + q x + r = 0$ $px^2+qx+r= 0$ can be re-arranged to ${(x + \frac{q}{2 p})}^{2} = - \frac{r}{p} + {(\frac{q}{2 p})}^{2}$ $(x+q/(2p))^2 = -r/p + (q/(2p))^2$ . The roots are in the form ' $a + i b$ $a+i b$ '

We had derived the complex number notation $a + i b$ $a+i b$ specifically for quadratic equations. Let us examine other cases where complex numbers are defined.

explaining the problem

There are $3$ $3$ possible solutions to the equation of degree $3$ $3$ . So, the equation $x^{3} = - 1$ $x^3 = -1$ has $3$ $3$ solutions. One of the solutions is $x = - 1$ $x=-1$ . We'll look into finding the other two solutions.

The irrational numbers are defined as numerical expressions of various kinds
• $\sqrt{2}$ $sqrt(2)$
• $\sqrt[3]{5}$ $root(3)(5)$
• $π$ $pi$
• $3 - 2 \sqrt[7]{5}$ $3-2root(7)(5)$

In case of complex numbers for defined quadratic equations, a new number was defined $i$ $i$ and the numbers are presented in a numerical expression form $a + i b$ $a+ib$ .

The questions is "Do we have to define a number of new elements (like $i = \sqrt{- 1}$ $i=sqrt(-1)$ ) for each of the following?"
• three solutions of $x^{3} = - 1$ $x^3=-1$
• four solutions of $x^{4} = - 1$ $x^4=-1$
• n solutions of $x^{n} = - 1$ $x^n=-1$

No, That is not necessary as the Euler's representation solves this problem and provides a generic form for all complex numbers.

3 solutions to x cube

Consider the example $x^{3} = - 1$ $x^3=-1$ . To find the three roots of this equation of degree $3$ $3$ we use $r e^{i θ}$ $r e^(i theta)$ form for the real number on the right-hand-side. `

The equation $x^{3} = - 1$ $x^3=-1$ is given as $x^{3} = 1 e^{i (2 n + 1) π}$ $x^3 = 1 e^(i (2n+1)pi)$ where $n = 0, 1, 2, 3, 4 ...$ $n=0,1,2,3,4...$

The solution is found by taking third root of right-hand-side and substituting $n = 0, 1, 2$ $n=0,1,2$ .

Solving that we get the three solutions
• $e^{\frac{i π}{3}}$ $e^((i pi)/3)$
• $e^{i π}$ $e^(i pi)$
• $e^{\frac{i 5 π}{3}}$ $e^((i 5pi)/3)$

For values $n \geq 3$ $n>=3$ , the solution is same as that of $n = 0, 1, 2$ $n=0,1,2$ . For example, $n = 3$ $n=3$ , we get $e^{\frac{i 7 π}{3}}$ $e^((i 7pi)/3)$ $= e^{\frac{i π}{3}}$ $= e^((i pi)/3)$ -- which is same as the solution for $n = 0$ $n=0$ .

generic form is defined

Generalizing what we have learned : For any algebraic expression with solution as a complex number, the solution can be equivalently given in the form $a + b i$ $a+bi$ where $i = \sqrt{- 1}$ $i=sqrt(-1)$ . This representation is named as 'complex number'.

The word "complex" means: consisting of many different parts.

A complex number $a + b i$ $a+bi$ is called complex as it has two parts to it
• $a$ $a$ - called the real part
• $b$ $b$ - called the imaginary part.

The word 'real' means: actually existing as a thing in reality.

The word 'imaginary' means: that is imagined and does not exist.

For a complex number $a + b i$ $a+bi$
• $a$ $a$ is the real part
• $b$ $b$ is the imaginary part.

Historically, the real number solutions for polynomials were easily found, and the rest of the solutions were named with meaning "imaginary". This name stuck.

Later in this course, in the page "Modeling Sine Waves using Complex Numbers" the significance of real and imaginary parts are explained.

Is $\sqrt{2} + i \sqrt[5]{7}$ $sqrt(2)+i root(5)(7)$ a complex number?

The answer is 'Yes. The real and imaginary parts can be real numbers.'. Irrational numbers are real numbers.

Is $\sqrt{- 2} + i 5$ $sqrt(-2)+i5$ a complex number?

The answer is 'Yes. It can be represented in $a + i b$ $a+ib$ form.' Simplify the same as $\sqrt{2} i + i 5$ $sqrt(2)i+i5$ which is $0 + i (5 + \sqrt{2})$ $0+i(5+sqrt(2))$ . The result is in $a + i b$ $a+ib$ form.

summary

Generic Form of Complex Numbers : Complex numbers are in the form $a + b i$ $a+bi$ where $a, b \in ℝ$ $a, b in RR$ and $i = \sqrt{- 1}$ $i=sqrt(-1)$ .

Outline

The outline of material to learn "complex numbers" is as follows.

Note : Click here for detailed overview of Complex-Numbers

→ Complex Numbers in Number System

→ Representation of Complex Number (incomplete)

→ Euler's Formula

→ Generic Form of Complex Numbers

→ Argand Plane & Polar form

→ Complex Number Arithmetic Applications

→ Understanding Complex Artithmetics

→ Addition & Subtraction

→ Multiplication, Conjugate, & Division

→ Exponents & Roots

→ Properties of Addition

→ Properties of Multiplication

→ Properties of Conjugate

→ Algebraic Identities