Complex Numbers

Simplifying Complex Numbers

In the radicals section, we noted that we won’t get a real number out of a square root of a negative number. For instance √ – 9 isn’t a real number since there is no real number that we can square and get a NEGATIVE 9.

Now we also saw that if a and b were both positive then √ ab = √ a · √ b . For a second let’s forget that restriction and do the following.

– 9 = √ (9)(– 1) = √ 9 · √ – 1 = 3√ – 1

Now, √– 1 is not a real number, but if you think about it we can do this for any square root of a negative number. For instance,

– 100 = √ 100 · √– 1 = 10 √– 1

– 5 = √ 5 · √– 1

– 290 = √ 290 · √– 1

So, even if the number isn’t a perfect square we can still always reduce the square root of a negative number down to the square root of a positive number (which we or a calculator can deal with) times √– 1.

If we just had a way to deal with √– 1 we could actually deal with square roots of negative numbers. Well the reality is that, at this level, there just isn’t any way to deal with √– 1.So instead of dealing with it, we will “make it go away” so to speak by using the following definition.

i = √ – 1

Note that if we square both sides of this we get,

i² = – 1

It will be important to remember this later on. This shows that, in some way, i is the only “number” that we can square and get a negative value.

Using this definition, all the square roots above become,

– 9 = 3i

– 100 = 10i

– 5 = √ 5 i

– 290 = √ 290 i

These are all examples of complex numbers.

The natural question at this point is probably just why do we care about this? The answer is that sometimes we will run across the square roots of negative numbers and we’re going to need a way to deal with them. So, to deal with them we will need to discuss complex numbers.

So, let’s start out with some of the basic definitions and terminology for complex numbers. The standard form of a complex number is

a + bi

where a and b are real numbers and they can be anything, positive, negative, zero, integers, fractions, decimals, it doesn’t matter. When in the standard form a is called the real part of the complex number and b is called the imaginary part of the complex number.

For example, in the complex number

3 + 2i

a – that is, 3 in the example, is the real part.

b – that is, 2 in the example, is the imaginary part.

Here are some examples of complex numbers.

3 + 2i 6 – 10i########## MISSING EQUATION ##############+ i16i113

The last two probably need a little more explanation. It is completely possible that a or b could be zero and so in 16i the real part is zero. When the real part is zero we often will call the complex number a purely imaginary number. In the last example (113 ) the imaginary part is zero and we actually have a real number. So, thinking of numbers in this light we can see that the real numbers are simply a subset of the complex numbers.

The conjugate of the complex number, a + bi,is the complex number, a – bi. In other words, it is the original complex number with the sign on the imaginary part changed. Here are some examples of complex numbers and their conjugates.

3 + \frac{1}{2}i
3 – \frac{1}{2}i
12 – 5i
12 + 5i
1 – i
1 + i
45i
– 45i
101

101

Notice that the conjugate of a real number is just itself with no changes.

Adding and Subtracting Complex Numbers

The commutative, associate and distributive properties for real numbers also apply to complex numbers. Adding, subtracting, multiplying, and dividing complex numbers is very similar to adding, subtracting, multiplying, and dividing binomials. When adding or subtracting, we add the real parts and add the imaginary parts.

The easiest way to think of adding and/or subtracting complex numbers is to think of each complex number as a polynomial and do the addition and subtraction in the same way that we add or subtract polynomials.

Example 1. Perform the indicated operation and write the answer in standard form. (4 + 12i) – (3 – 15i)

Solution.

(4 + 12i) – (3 – 15i)

= 4 + 12i – 3 + 15i

= 1 + 27i

Multiplying Complex Numbers

Next let’s take a look at multiplication. Again, with one small difference, it’s probably easier to just think of the complex numbers as polynomials and multiply them out as you would multiply polynomials. The one difference will come in the final step.

Example 2. Multiply the following and write the answer in standard form.

(4 + i)(2 + 3i)

Solution.

(4 + i)(2 + 3i)

= 8 + 12i + 2i + 3i²

= 8 + 14i + 3(–1)

= 5 + 14i

Dividing Complex Numbers

The Product of Complex Conjugates shows that when two complex conjugates are multiplied, the product is a real number.This is a nice general formula that will be convenient when it comes to division of complex numbers.

(a + bi)(a bi)

= a² – abi + abi – b²i²

= a² + b²

So, when we multiply a complex number by its conjugate we get a real number given by,

Product of Complex Conjugates

(a + bi)(a bi) = a² + b²

Now, we gave this formula with the comment that it will be convenient when it came to dividing complex numbers.

Example 3. Write the following in standard form.

\frac{3 - i}{2 + 7i}

Solution.

Standard form does not allow for any i's to be in the denominator. So, we need to get the i's out of the denominator.

This is actually fairly simple if we recall that a complex number times its conjugate is a real number. So, if we multiply the numerator and denominator by the conjugate of the denominator we will be able to eliminate the i from the denominator.

Now that we’ve figured out how to do these, let’s go ahead and work the problems.

 \frac{3 - i}{2 + 7i} = \frac{(3 - i)}{(2 + 7i)} \cdot \frac{(2 - 7i)}{(2 - 7i)} = \frac{6 - 23i + 7i^2}{2^2 + 7^2} = \frac{-1 - 23i}{53} = \frac{1}{53} - \frac{23}{53}i

Notice that to officially put the answer in standard form we broke up the fraction into the real and imaginary parts.

Evaluate: i21
Multiply (6-4i)(2+4i)
Add: (3+2i) + (4+5i)