Imaginary and Complex Numbers

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Mathematicians being what they are, were frustrated by the fact that they couldn't solve every quadratic equation in the universe since some of the solutions involved the square root of a negative number and such numbers did not exist in the set of Reals. This frustration led them to create two new sets of numbers called the imaginary numbers and the complex numbers.

What they did was define i to be the square root of 1, and since any negative number n can be expressed as 1 times n , if we need to express the square root of 9, we write it as root 9 times root of 1. In the set of imaginary numbers, the square root of 9 = 3i as it's 3 times the square root of 1.

Remember, when we solved quadratic equations with the quadratic formula, there were times that we had to answer "there are no REAL roots or solutions for this equation" because the
discriminant (b² 4ac) of the quadratic formula was negative or less than zero. Once we introduce the imaginary and complex numbers, every quadratic equation has a solution. Now, we have to state the nature of the solutions as well as the number of solutions. This lesson will cover the basic definitions, properties and calculations of the imaginary and complex numbers, the next one will cover solving different types of equations with imaginary and complex numbers.

Definition of i

The constant i is very much like the constants e, the base of the natural logs; and o, the ratio of Circumference to diameter in a given circle, in that it is defined as a precise, and unique value.
It is unlike e, and o, because those 2 are real whereas i exists only in the human mind. In other words, we could create a line or piece of string that is exactly o units in length, we can locate it precisely on a number line through a geometric construction, however we cannot do any of those things with i. It is imaginary, so we can only imagine it. We can however use it to define polar coordinates and other mathematical calculations.

 The constant i is defined as the square root of - 1.This means that i² = –1

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An imaginary number is one which can be expressed as a nonzero real number times i.
Say we're solving a quadratic equation with the quadratic formula and we get:

If we were asked to solve the equation in the set of Real numbers, there would be no solutions.
However, we can solve this equation in the set of imaginary numbers. Here's how to proceed.

This quadratic equation has 2 solutions in the set of Imaginary numbers.

 An Imaginary number is of the form bi whereb is a nonzero Real number,

Examples

Express these roots in the form bi.

 a) b) c) d)

Powers of i

By definition, we can see that i² = 1, i³ = i² % i = 1 % i = i and i 4 = (i²)² = (1)² = + 1.

Since i 5 = i ³% i ², it equals ( i) % ( 1) = i.

This shows that the powers of i cycle through 4 values: i, 1, i, and + 1.

So to simplify or evaluate powers of i, we express i n as (i²)n / 2 (i) k .
Then n / 2 and k will determine where we are in the cycle of values for i.
Even powers of i will be either 1 or + 1. Odd powers of i will be either i or i.

Examples

Evaluate these powers of i.

 a) i 37 = (i ²) 18 i = (–1)18 i = i b) i 58 = (i ²) 29 = (– 1) 29 = – 1 c) i 40 = (i ²) 20 = (– 1)20 = + 1

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 A Complex number is of the form a + bi wherea and b are Real numbers.A Complex number has a Real part and an Imaginary part. The Real part is a and the Imaginary part is bi.

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These numbers behave exactly like the Reals do in that operations on them have the same properties as operations in the set of Real numbers. Addition and multiplication are commutative. Multiplication can be distributed on a sum or difference etc. So all we have to do to simplify these imaginary numbers is perform the usual algebraic operations, collect like terms, and then replace any powers of i with their proper values.

Examples

a) (8 + 7i) + (9 5i) = 8 + 9 + 7i 5i = 17 + 2i.

b) (5 + 3i) (2 4i ) = 10 + 6i 20i 12i ² = 10 14i 12(1) = 22 14i.

c) 5i (4 3i) = 20i + 15i ² = 20i + 15(1) = 20i 15 or (15 + 20i).

Conjugates and Division of Complex Numbers

To Divide by a complex number we use a technique almost identical to the one we use when we rationalize the denominator of a fraction with radicals, (square roots). In lesson 3 of the Algebra MathRoom, we learn how to rationalize a binomial denominator with roots. We multiply our fraction by a unit fraction with the conjugate surd in both numerator and denominator. We do this because we know that (a + b)(a b) = a² b², and the square of a square root is a Real number.

For example, say we have to rationalize:

We would proceed like this:

With Complex numbers, we shoot for (b i) ² because i ² = 1 so ( b)² = a² + b²,
a Real number with no i.

The conjugate of a + bi is a bi.

(a + bi )(a bi ) = (b² i ²) = a² + b²

So, to change a complex fraction into a Real number, multiply it by a unit fraction c / c where c equals the conjugate of the fraction's denominator.

Note also that in the set of Complex Numbers, a sum of squares can be factored!

So now we can say that a² + b² = (a + bi )(a bi )

Example

Rewrite this quotient as a single Complex number

We will multiply by a unit fraction c / c with c = (1 4i ) , like this:

Now we have a Complex number of the form a + bi.

Note, for calculation purposes we could write it as

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Open solutions in a New Window.

1) Simplify and leave answers in terms of i.

 a) b) c) d) e) f) g) h) i) j)5i 39 = k) – 4i 46 = l) – 7i 19 =

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 a) (2 + 3i) + (4 + 2i) b) (5 – 2i) – (6 + 3i) c) (3 – 7i) – (5 – 4i) d) (3 – 7i)(3 + 7i) e) (1 + 2i)(1 + 3i) f) 5i (3 – 4i) g) (2 + 3i)² h) (3 – 2i)²

3) Write each quotient as a Complex number.

 a) b) c) d)

4) Factor these sums of squares:

 a) 4x² + 25y² b) 81 + 16m²

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1) Simplify and leave answers in terms of i.

 a) b) c) d) e) 5i + 4i = 9i f) 6i – 2i = 4i g) h) i) j)5i 39 = 5(i²) 19 i = – 5i k) – 4i 46 = – 4(i²) 23 = 4 l) –7i 19 = –7 (i²) 9 i = 7i

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 a) (2 + 3i) + (4 + 2i) = = 6 + 5i b) (5 – 2i) – (6 + 3i) = –1 – 5i c) (3 – 7i) – (5 – 4i) = –2 – 3i d) (3 – 7i)(3 + 7i) 9 + 49 = 58 e) (1 + 2i)(1 + 3i) = 1 + 5i + 6i² = –5 + 5i f) 5i (3 – 4i) = 15i + 20 g) (2 + 3i)² = 4 + 12i + 9i² = –5 + 12i h) (3 – 2i)² = 5 – 12i

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3) Write each quotient as a Complex number.

 a) b) c) d)

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4) Factor these sums of squares:

 a) 4x² + 25y² = (2x + 5yi)(2x – 5yi) b) 81 + 16m² = (9 + 4mi)(9 – 4mi)

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