# Imaginary (Complex) Numbers

This section covers:

# Introduction to Imaginary Numbers

Think of imaginary numbers as numbers that are typically used in mathematical computations to get to/from “real” numbers (because they are more easily used in advanced computations), but really don’t exist in life as we know it.  Yet they are real in the sense that they do exist and can be explained quite easily in terms of math – think  of(infinity) – does that really exist?.  And we’ve all had imaginary friends, right?

Actually, imaginary numbers are used quite frequently in engineering and physics, such as an alternating current in electrical engineering, which is usually represented by a complex number.  (Don’t worry; I don’t know what an alternating current is, either.)

A complex number consists of a “real” part and an “imaginary” part, and typically  looks like , where “a” is the real part, and “b” is the imaginary part, following by “i”, to indicate the “imaginary” unit.  For now, let’s just learn about the imaginary part.   (Note that all real numbers, even without the imaginary part, are considered to be complex, as we saw in Types of Numbers).

Let’s look at the following graphs and notice that the parabolas never touch the x axis, so there aren’t any x-intercepts, although the “roots”, “solutions”, and “values” are “complex” or “imaginary”.

As we learned earlier, since these graphs have no roots, they have a discriminant that is less than 0.  The graphs will never touch the x axis, yet we can still find imaginary roots, and the roots will have “i”s in them, as we see later.   But we can’t find these roots with a graphing calculator!

# Working with “i”,  the

Before working problems that have imaginary solutions, we need to learn about the value of a special “number” called “i”.   i is simply , which can’t exist in our “real” system, since we can never take two “real” numbers multiplied together to get  –1.  Since i equals , then it follows that:

So,   Similarly,  (if you put the i at the end, make sure it is clearly outside of the square root sign in this case).

And here’s something really cool: when we multiply i’s together, we notice a pattern:

Note that there’s a repeating pattern when raising “i” to an exponent. Notice that every fourth exponent number repeats, so , and so on.  Because of this, we can easily compute “i” raised to any exponent by dividing that exponent by four, and examining the remainder, as shown in the examples:

Again, when dealing with complex numbers, expressions contain a real part and an imaginary part.  Together they form a complex number that typically looks like a + bi, where “a” is the real part, and “b” is the imaginary part, following by “i”, to indicate the “imaginary” unit.  (Later, in Pre-Calculus, we’ll see how these can be graphed on a coordinate system, where the “x” is the real part and the “y” is the imaginary part.)

For example, “4 + 5i” indicates the number , and we cannot mix the real parts with the imaginary parts when adding or subtracting, so that the “i’s” are treated somewhat like variables (like radicals were thought as variables, back in the Exponents and Radicals in Algebra section).

So when we perform operations on i, we pretty much treat it like a variable, except when we’re multiplying the “i’s” together – and then we can simplify.   Note that for good “math grammar” we want our final answer to be in the form a + bi.  Here are some examples:

You can also put complex expressions in the graphing calculator:

(Note that the complex conjugate that we used to simplify a denominator with an imaginary number in it is similar to the radical conjugate we learned about here in the Introduction to Quadratics section.)

# Quadratic Formula with Complex Solutions

Now let’s solve a quadratic equation that has complex (imaginary) solutions.

Let’s take the equation .  We know that since the discriminant  for a, b, and c in  is negative ( –4), there are no real solutions to the equation.

Let’s use the quadratic equation to find this solution, and one that’s a little more complicated:

# Completing the Square with Complex Solutions

Let’s try completing the square with a quadratic with complex solutions:

Yeah!  We got the same answers as when we solved with the Quadratic Equation!

We learned earlier here in the Introduction to Quadratics section that when we have an irrational value for a root, the conjugate is also a root.  (For example, if  is a root, then  is also a root).

Similarly, if we have a complex root, the complex conjugate is also a root; this is called the Complex Conjugate Root Theorem, or Complex Conjugate Zeros Theorem.  For example, if  3 + i is a root, then 3 – i is also a root.  Interesting!

Learn these rules and practice, practice, practice!

Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software.  You can even get math worksheets.

You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions.  There is even a Mathway App for your mobile device.  Enjoy!

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On to Compositions of Functions, Even and Odd, and Increasing and Decreasing – you are ready!

## 18 thoughts on “Imaginary (Complex) Numbers”

• Maureen,
Thanks so much for your nice comment! This makes me want to keep writing – I have a lot more to write.
Let me know if you have any question in math 🙂
Lisa

• Please can you tell me if a radical equation can have roots which are complex numbers?

5*SQRT(36+x^2) + 4*(20-x)=0

This gives a quadratic equation: 9*x^2 + 640*x – 5500 = 0
whose roots are +7.75 and -78.86. Checking back sees that neither of these roots fits the original equation. So everything I read says “no solution”. Does this mean “no REAL number solution”? Can there be a solution that is a complex number, and if so, how can I find it?

Also, why does the part under the square root sign (36+x^2) have to be positive and real? If x were a complex number, I know how to find the square root of a complex number (polar co-ordinates).

I was thinking that if x was a complex number, then SQRT(36+x^2) would be a complex number, and also -4x would be complex, and perhaps the non-real part would be cancelled out, so that the equation could work.

Hope this is clear. Thanks.

• Thanks for writing! This is a GREAT question and I think it’s beyond my abilities right now. I did get the same quadratic you did, and even graphed it, and found there were no real solutions. I’m just not sure if there are complex solutions, and if so, how to find them. Did you find out anything more? I also sent the problem off to a math genius to see if she can solve it – I’ll let you know. Lisa

• Hi Lisa,

Many thanks for taking the time to look at my question. I have spent several more hours on this (!) and gone up a few blind alleys, but am no further forward.

If you hear back from your colleague please let me know what she says.

Thanks
Sue

• I think I have an answer – there are no complex roots (thus no solutions). Here’s what I got from my friend:
Remember that the fundamental theorem of algebra is for polynomial equations of degree >= 1 (Fundamental Theorem of Algebra says that (from Wikipedia) every non-constant single-variable polynomial with complex coefficients has at least one complex root.)
Polynomials have 1 or more monomials. Therefore, the fundamental theorem of algebra does not apply to radical equations because the exponents are not integers, they are fractions.
With radical equations, we typically isolate the radical term and square both sides. This may introduce extraneous solutions so you must check your solutions in the ORIGINAL equation. Some or all of your solutions may be extraneous.
Also, using common sense is helpful. In the radical problem below, you might wonder if there is an imaginary solution. If X is imaginary, then the x^2 inside the radical would just be a negative integer. If it were 6i then the first part would reduce to zero. If less than 6i it would reduce to a positive irrational number and the second part would be imaginary and they would NOT cancel each other.
After isolating the radical and squaring both sides, rearranging to get a quadratic equation, and solving. … You will find 2 extraneous solutions. Neither makes the ORIGINAL equation true. Therefore, there is no solution!
And here’s a video of the problem: https://m.youtube.com/watch?v=Z2lrX6vhZRY

• Thanks for your reply. Unfortunately I’m still a bit confused. What would happen if X > 6i ? Say X = 10i, then X^2 = -100, Z = -64 (where Z = 36+X^2) and the square root of Z = 8i. Then 5Z=40i and 4X=40i, so this is what I meant when I said they could cancel each other out. Of course, this doesn’t take into account the constant 80, but I was hoping to find a value for X=a+bi which would solve this.

I see that in the video that X>20, which means that the square root of Z is real and positive. But what if it isn’t real? Please have a look at this example below, hopefully this will explain what I’m trying to do.

5 * sqrt(36+x^2) – (4x+3) = 0

25 * (36+x^2) = 16x^2 + 24x + 9

900 + 25x^2 = 16x^2 + 24x + 9

9x^2 – 24x + 891 = 0

a=9, b=-24, c=891

b^2 – 4ac = -31,500 and sqrt(b^2 – 4ac) = (+ or -) i 177.482394

x = 1.333333 + i 9.860133 and x = 1.333333 – i 9.860133

Both values of x fit the original equation (see below for proof). Which means that the argument of the radical, Z, is neither positive nor real.

In this example it was easy to find the complex value for X because it is the result of the quadratic equation. I don’t think there is a solution for my original equation
[5 * sqrt(36+x^2) – 4 * (20-x) = 0] but I don’t know how to prove it, given that X might equal a+bi and the argument of the radical might be negative.

Hope this makes sense. Thanks for your time.

Proof:

x = 1.333333 + i 9.860133 (also works for x = 1.333333 – i 9.860133)

x^2 = -95.444445 + i 26.293681

Z = 36 + x^2 = -59.444445 + i 26.293681

sqrt(Z) = 1.666627 + i 7.8881 (you can square this to check it equals Z)

5 * sqrt (Z) = 8.333135 + i 39.4405

4x + 3 = 8.333333 + i 39.4405

5 * sqrt (Z) – (4x + 3) = 0 (allowing for rounding errors)

• I totally hear you! I’m sending your response to my colleague who tried to do the problem. Here’s another response I got for this problem – does this help? Lisa
—————————————————
To answer your question above, it does depends on the domain of your problem. For example when the domain is the reals (no imaginary numbers allowed), then it is possible to for no solution to exist. (Clearly, there is no real-valued solution exists in the equation x^2 = -1.)

Now let’s suppose our domain is the complex plane. Then we have to use the more generalized definitions for mathematical operations such as addition, multiplication, etc. Depending the complex notation (phase and angle or real and imaginary), either addition or multiplication (respectively) are fairly difficult to wield computationally (by hand), so I’d rather not work it out that way. It is possible, however, for no complex solution to exist. (Also, that’s what the sophisticated calculator on my computer is saying.)

To see this, consider sqrt(x^2 +1) – sqrt(x^2 – 1). In order for this equation to hold, x^2 + 1 = x^2 – 1, or 1 = -1, which has no solution.

• Thanks for writing! This is tricky – you have to turn the numbers into complex numbers first, and then multiply: 2i x 3i = 6i^2, so -6. Does that make sense? Lisa

• Thanks for writing! The reason complex solutions aren’t thought of as x-intercepts is because the solutions don’t lie on the x-axis, like real solutions do. Think of a cup extended above the x-axis and the lowest point of the cup never touches it. That’s what a complex solution looks like (or it could be an upside cup below the x-axis that isn’t is high as it). Does that make sense? Lisa

1. How can we solve for the question like this?
(2-i)x + 3y=6-2i, find x and y when x and y are complex number with x being conjugate of y? Hope you can help me.(

• Thanks for writing! Here’s what I got on this problem: I let x = a + bi and y = a – bi (conjugate). Then I got (2-i)(a+bi)+3(a-bi)=6-2i. Then I multiplied out to get 2a-ai+2bi-bi^2=6-2i. Then I simplified to get 5a+b+i(-a-b)=6-2i (remember that i^2=-1). Then I got 5a+b=6 and -a-b=-2. Then I solved the system to get a=1 and b=1. Then x=1+i and y=1-i. Does that make sense? Lisa

• Okay . Thank you very much. I want to ask, what means that “x being conjugate of y”? Is it x being conjugate or y being conjugate? If x or y being conjugate, then we need to conjugate the final answer of x//y or not?

• Good question! x being conjugate of y means the same as y being conjugate of x. This is because, with conjugates, one is a + bi, and the other is a – bi. So the conjugate goes either way. Does that make sense? Lisa

2. Thanks a lot lisa,
i just started doing my degree n lost all my school notes!this really helps me.+ well organised way of writing its very easy to learn 😀 thanks a lot -chirantha