# Complex Numbers

A **complex number** is a number that can be written in the form \( a + bi \), where \( a\) and \( b\) are real numbers and \( i \) is the imaginary unit defined by \( i^2 = -1 \).

The set of all complex numbers is often denoted by \( \mathbb{C} \).

Note that the complex numbers include all the real numbers (with \( b=0 \) in \(a+bi\) above).

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## Graphical Interpretation

Any complex number \( a + bi \) with \( b \neq 0 \) will have an imaginary part and thus will not be found on the real number line. As a result, it is helpful in many circumstances to think of the complex numbers as residing on the Complex Plane, where the imaginary numbers make a new axis perpendicular to the real numbers.

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Any complex number can thus be represented by an ordered pair \( (a,b) \) on the complex plane.

## Complex Numbers Arithmetic

When adding complex numbers together, the basic algebraic principle of combining like terms is extremely useful.

Addition of Complex Numbers:Given complex numbers \( a + bi \) and \( c+di \), their sum is

\[ (a+c) + (b+d)i.\ _\square \]

## What is \((4+3i)+(2+2i)\)?

Adding the real and imaginary parts separately, we get:

\[ 4+3i+2+2i = (4+2) + (3+2)i = 6+5i.\ _\square\]

Notice how real numbers were only added to other real numbers and imaginary numbers were only added to other imaginary numbers.

A few additional examples:

- \((3-4i)+(3+2i) = 6-2i\)
- \((4-2i)-(-2-5i) = 6+3i\)
- \((3i)+(3+5i) = 3+8i.\)

When multiplying two complex numbers together, we must remember both the distributive property and the definition of the imaginary unit.

Multiplication of Complex Numbers:Given two complex numbers \( a + bi \) and \( c+di \), their product is

\[ \begin{align} (a+bi) \times (c+di) &= a(c+di) + bi(c+di) \\ &= (ac) + (ad)i + (bc)i + (bd)i^2 \\ &= (ac) + (ad+bc)i + (bd)(-1) \\ &= (ac - bd) + (ad+bc)i .\ _\square \end{align} \]

## What is \((3+2i)(4-2i)\)?

:Solution 1

By the definition above, \( a=3, b=2, c=4, d=-2\), and so the product is\[ (12 - (-4)) + (-6+8)i = 16 + 2i.\ _\square \]

:Solution 2

Working out the product from first principles, we have\[ \begin{align} (3+2i)(4-2i) &= 3(4-2i)+2i(4-2i) \\ &= 12-6i + 8i -4i^2 \\ &= 16 +2i.\ _\square \end{align} \]

## Complex Roots

By the Fundamental Theorem of Algebra every polynomial of degree \( n\) has exactly \( n\) roots. However, sometimes these roots are complex numbers rather than real numbers. As a simple example, \( x^2 +1 = 0 \) is only true when \( x^2 = -1 \), which of course is only possible if \( x =\pm i \).

## Find the roots of the equation \(2x^2 + 1 = 0.\)

Subtracting \(1\) and then halving both sides, we get \(x^2 = \dfrac{-1}{2}.\) Thus,

\[\begin{align} x &= \pm \sqrt \frac{-1}{2}\\ &=\pm \sqrt{-1} \sqrt\frac{1}{2}\\ &=\pm i \sqrt{\frac{1}{2}}\\ &=\pm \frac{i}{\sqrt{2}} .\ _\square \end{align}\]

## Factorize the following quadratic: \(x^2 + 6x + 10 = 0.\)

Computing the discriminant \(D\), we get

\[D = b^2 - 4ac = 6^2 - 4\times1\times10 = -4 .\]

It is less than \(0\) and thus we can conclude that the quadratic has a pair of complex roots with imaginary components. To find them, we use the quadratic formula as follows:

\[\begin{align} x &= \frac{ -b \pm \sqrt{D}}{2a} \\ & = \frac{-6 \pm \sqrt{-4}}{ 2 \times 1} \\ & = \frac{ -6 \pm 2\sqrt{-1}}{2} \\ &=-3 \pm i . \end{align}\]

Factorizing the expression, we get

\[\begin{align} x^2+6x+10 &=(x-(-3+i))(x-(-3-i))\\ &=(x+3-i)(x+3+i). \ _\square \end{align}\]

## Find the roots of \(3x^2 + 5ix + 7 = 0.\)

Computing the discriminant \(D\), we get

\[D = b^2 - 4ac = (5i)^2 - 4\times3\times7 = -109. \]

It is indeed less \(0\) and thus we can conclude that the quadratic has a pair of complex roots. To find them, we use the quadratic formula as follows:

\[\begin{align} x &= \frac{ -b \pm \sqrt{D}}{2a} \\ &= \frac{-5i \pm \sqrt{-109}}{ 2 \times 3} \\ &= \frac{ -5i \pm \sqrt{109} \sqrt{-1}}{6}. \end{align} \]

Thus, the roots are

\[\begin{array} &x =\left (\frac{-5-\sqrt{109}}{6}\right)i &\text{and} &x = \left(\frac{-5+\sqrt{109}}{6}\right)i .\end{array} \ _\square\]

## Problem Solving - Basic

## What is \( i^{201} \)?

Since \( i^2 = -1 \) by definition, and thus \( i^4 = i^2 \cdot i^2 = (-1)(-1) = 1 \), we can express \( i^{201} \) as follows:

\[ i^{201} = i^{200}i = \left( \left(i^4\right)^{50} \right)i= 1^{50}i = i.\ _\square \]

## Problem Solving - Advanced

## If \( x = 1+2i \), what is the value of \( x^3 + 2x^2 + 4x + 25 \)?

:Solution 1

We could evaluate each of the terms individually and then add:

- \( x^3 = (1+2i)^3 = -11 - 2i \)
- \( 2x^2 = 2(1+2i)^2 = 2(-3+4i) = -6+8i \)
- \( 4x = 4(1+2i) = 4 + 8i \)
Therefore, \( x^3 + 2x^2 + 4x + 25 = (-11 - 2i)+( -6+8i) + (4 +8i) + 25 = 12+14i \). \(_\square\)

:Solution 2

Since \( x = 1 +2i \), \( x-1 = 2i \). Squaring both sides gives\[ x^2 -2x +1=4i^2 \Rightarrow x^2-2x +5 = 0. \qquad (1) \]

Then

\[ \begin{align} x^3 + 2x^2 + 4x +25 &= \left(x^2-2x+5\right)\left(x+4\right) + 7x + 5 \\ &= 0 \cdot (x+4) + 7x +5 \qquad (\text{by }(1)) \\ &= 7(1+2i) + 5 \\ &= 12 + 14i. \ _\square \end{align} \]