Mathematics Magazine for Grades 1-12

##### Theory:   Complex Numbers

Why do we need new numbers?

The hardest thing about working with complex numbers is understanding why you might want to. Before introducing complex numbers, let's backup and look at simpler examples of the need to deal with new numbers.

If you are like most people, initially number meant whole number, 0,1,2,3,... Whole numbers make sense. They provide a way to answer questions of the form "How many ... ?" You also learned about the operations of addition and subtraction, and you found that while subtraction is a perfectly good operation, some subtraction problems, like 3 - 5, don't have answers if we only work with whole numbers. Then you find that if you are willing to work with integers, ...,-2, -1, 0, 1, 2, ..., then all subtraction problems do have answers!

Furthermore, by considering examples such as temperature scales, you see that negative numbers often make sense.

Now we have fixed subtraction we will deal with division. Some, in fact most, division problems do not have answers that are integers. For example, 3 ÷ 2 is not an integer. We need new numbers!

Now we have rational numbers (fractions).

There is more to this story. There are problems with square roots and other operations, but we will not get into that here. The point is that you have had to expand your idea of number on several occasions, and now we are going to do that again.

The "problem" that leads to complex numbers concerns solutions of equations.

Equation 1:  x2 - 1 = 0.

Equation 1 has two solutions, x = -1 and x = 1. We know that solving an equation in x is equivalent to finding the x-intercepts of a graph; and, the graph of y = x2 - 1 crosses the x-axis at (-1,0) and (1,0).

Equation 2x2 + 1 = 0

Equation 2 has no solutions, and we can see this by looking at the graph of y = x2 + 1.

Since the graph has no x-intercepts, the equation has no solutions. When we define complex numbers, equation 2 will have two solutions.

The Number i

Consider Equations 1 and 2 again.

 Equation 1 Equation 2 x2 - 1 = 0. x2 + 1 = 0. x2 = 1. x2 = -1.

Equation 1 has solutions because the number 1 has two square roots, 1 and -1. Equation 2 has no solutions because -1 does not have a square root. In other words, there is no number such that if we multiply it by itself we get -1. If Equation 2 is to be given solutions, then we must create a square root of -1.

Definition: The imaginary unit i is defined by

The definition of i tells us that i2 = -1. We can use this fact to find other powers of i.

Example 1.

i3 = i2  i = -1 i = -i.

i4 = i2  i2 = (-1)  (-1) = 1.

Exercise 1:

Simplify i8 and i11.

We treat i like other numbers in that we can multiply it by numbers, we can add it to other numbers, etc. The difference is that many of these quantities cannot be simplified to a pure real number.

For example, 3i just means 3 times i, but we cannot rewrite this product in a simpler form, because it is not a real number. The quantity 5 + 3i also cannot be simplified to a real number.

However, (-i)2 can be simplified. (-i)2 = (-1 i)2 = (-1)2  i2 = 1  (-1) = -1.

Because i2 and (-i)2 are both equal to -1, they are both solutions for Equation 2 above.

The Complex Plane

Definition: A complex number is one of the form a + bi, where a and b are real numbers. a is called the real part of the complex number, and b is called the imaginary part.

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. I.e., a + bi = c + di if and only if a = c, and b = d.

Example

2 - 5i.

6 + 4i.

0 + 2i = 2i.

4 + 0i = 4.

The last example above illustrates the fact that every real number is a complex number (with imaginary part 0). Another example: the real number -3.87 is equal to the complex number -3.87 + 0i.

It is often useful to think of real numbers as points on a number line. For example, you can define the order relation c < d, where c and d are real numbers, by saying that it means c is to the left of d on the number line.

We can visualize complex numbers by associating them with points in the plane. We do this by letting the number a + bi correspond to the point (a,b).

Complex Arithmetic

When a number system is extended the arithmetic operations must be defined for the new numbers, and the important properties of the operations should still hold. For example, addition of whole numbers is commutative. This means that we can change the order in which two whole numbers are added and the sum is the same: 3 + 5 = 8 and 5 + 3 = 8.

We need to define the four arithmetic operations on complex numbers.

To add or subtract two complex numbers, you add or subtract the real parts and the imaginary parts.

 (a + bi) + (c + di) = (a + c) + (b + d)i. (a + bi) - (c + di) = (a - c) + (b - d)i.

Example

(3 - 5i) + (6 + 7i) = (3 + 6) + (-5 + 7)i = 9 + 2i.

(3 - 5i) - (6 + 7i) = (3 - 6) + (-5 - 7)i = -3 - 12i.

Note

These operations are the same as combining similar terms in expressions that have a variable. For example, if we were to simplify the expression (3 - 5x) + (6 + 7x) by combining similar terms, then the constants 3 and 6 would be combined, and the terms -5x and 7x would be combined to yield 9 + 2x.

Multiplication

The formula for multiplying two complex numbers is

(a + bi)  (c + di) = (ac - bd) + (ad + bc)i.

You do not have to memorize this formula, because you can arrive at the same result by treating the complex numbers like expressions with a variable, multiply them as usual, then simplify. The only difference is that powers of i do simplify, while powers of x do not.

Example

 (2 + 3i)(4 + 7i) = 2 4 + 2 7i + 4 3i + 3 7 i2 = 8 + 14i + 12i + 21 (-1) = (8 - 21) + (14 + 12)i = -13 + 26i.

Notice that in the second line of the example, the i2 has been replaced by -1.

Using the formula for multiplication, we would have gone directly to the third line.

Division

Definition: The conjugate (or complex conjugate) of the complex number a + bi is

a - bi.

Conjugates are important because of the fact that a complex number times its conjugate is real; i.e., its imaginary part is zero.

(a + bi)(a - bi) = (a2 + b2) + 0i = a2 + b2.

Example

 Number Conjugate Product 2 + 3i 2 - 3i 4 + 9 = 13 3 - 5i 3 + 5i 9 + 25 = 34 4i -4i 16

Suppose we want to do the division problem (3 + 2i) ÷ (2 + 5i). First, we want to rewrite this as a fractional expression .

Even though we have not defined division, it must satisfy the properties of ordinary division. So, a number divided by itself will be 1, where 1 is the multiplicative identity; i.e., 1 times any number is that number.

So, when we multiply by , we are multiplying by 1 and the number is not changed.

Notice that the quotient on the right consists of the conjugate of the denominator over itself. This choice was made so that when we multiply the two denominators, the result is a real number. Here is the complete division problem, with the result written in standard form.

##### Solutions from the Previous Issue:

1.        Find all solutions of the equation 2 sin(2A) + 1= 0

Solution:

Let u = 2A; the equation is then equivalent to sin (u) = - 1/2, for which the solutions are:

u = 2A = 7π/6 + 2n π  and u = 2A = 11 π/6 + 2nπ, n = 0, 1, 2..

Hence the solutions for A are:

A = 7π/12 + n π and 11 π/12 +n π, n = 0, 1, 2..

2.        Find all solutions of the equation 3 tan3A- tan A = 0

Solution:

The left hand side of the equation can be factored as:

3 tan3A - tan A = tan A (3tan2A - 1) = 0

hence either tan A = 0  or tan A = . For A ,

tan A = 0  => A =0, π or 2 π

while A =  => A = π/6 or 5 π/6

and A =  => A = 5 π/6 or 11π/6

The solution set of the original equation is then

{0, π/6,  π, 5π/6, 7π/6, 11π/6, 2 π}

3.        Find all solutions of the equation cos (4A) = ½ in the interval [0 π].

Solution:

We first find all solutions of the related equation cos t =1/2 , and find that

t= π/3 + 2n π and t = - π/3 + 2n π, n = 0, 1, 2..

comprise all the solutions. Hence

4A = π/3 + 2n π or – π/3 + 2n π, n = 0, 1, 2..

from which it follows that

A= π/12 + nπ/2 or – π/3 + n π/2, n = 0, 1, 2..

The solution angles A which lie in the interval [0, 2 π] are therefore

{ π/12, 5π/12, 7 π/12, 11 π/12, 13 π/12, 17 π/12, 19 π/12, 23 π/12}

4.        Solve the equation 2cos2x + 6 cos x –3 = 0. Restrict solutions to the interval [0, 2 π).

Solution:

The substitution u = cos x yields the equation2u2 + 6u – 3 = 0, which is quadratic in u. We use the quadratic formula to solve for u= cosx:

u = cos x = = 0.4365 or –3.4365

If cos x = 0.4365  the calculator gives x =1.1191 as the acute solution, so we deduce that the other solution is x = 2 π – 0.4365 = 5.1641. The equation cos x = -3.4365 has no solutions, since| cos x | ≤ 1  for all x. The solution set is therefore

{0.4365, 5.1641}.

Proposed Problems:

1.       Find the terms a1, a3, a5, a6 for the next arithmetic progression:

a1, -7, a3, -1, a5, a6, 8, ... .

2.        A sequences is given such that  and  ; evaluate .

3.        If the sum of the first and the fifth term of an arithmetic progression is

equal to 10 and the product of the third term and the fifth term is equal

45 find the sum of the first twenty term of the progression.

4.        Solve the equation , knowing that its roots are in a geometric progression.

5.        Find the terms a2, a3, a5, a6 of the geometric progression:

, a2, a3, , a5, a6, , ... .

..read more on the written publication. Subscribe to the Mathematics Magazine for Grades 1-12