Complex Numbers - Online Tutor, Practice Problems & Exam Prep

Complex numbers, expressed in standard form as $a+bi$, consist of a real part $a$ and an imaginary part $b$. Operations like addition, subtraction, and multiplication follow familiar algebraic rules, while division requires the use of the complex conjugate to eliminate the imaginary unit from the denominator. Understanding these concepts is crucial for manipulating complex numbers effectively in various mathematical contexts.

Introduction to Complex Numbers

So we've worked with real numbers like 3 and imaginary numbers like \(2i\) separately, but you're often going to see expressions that have these two numbers together, something like \(3+2i\). These are called complex numbers when a real number and an imaginary number are added together. Now, complex numbers are going to be really important for us throughout this course and have a ton of different uses. While they might sound a little complicated at first, I'm going to walk you through what they are and how we use them step by step. So let's get started.

A complex number has a standard form of \(a+bi\). In this complex number, \(a+bi\), \(a\) is called the real part of the number, and \(b\) is called the imaginary part because it's multiplying \(i\), our imaginary unit. Now, it's important to know that \(b\) by itself is the imaginary part. It's not the whole term \(bi\). When I'm identifying the real and the imaginary part in the number I have up here, this \(3+2i\), \(3\) is going to be the real part.

Let’s look at a couple of different examples of complex numbers and identify the real and imaginary parts of each of them. So first, I have this \(4 - 3i\). Looking at this number, the real part \(a\) is going to be this \(4\) because it's out there by itself. It's not multiplying my imaginary unit. This is going to be my real part \(a\). Then \(b\), so I want to look for what is multiplying \(i\), my imaginary unit. In this case, it is negative 3. Now, it's important to look at everything that's multiplying our imaginary unit. So if it's a negative number, if it's a square root, if it's a combination of a number and a square root, I want to get everything that's multiplying my imaginary unit. So in this case, \(b\) is going to be a negative 3. That's what's multiplying my imaginary unit \(i\).

Let's look at another example. So here I have \(0+7i\). Now, if I look for the real part of my number, what is not multiplying my imaginary unit, I have this as \(0\). So that means that, \(a\), my real part is going to be \(0\). Then, \(b\), my imaginary part, the part that's multiplying my imaginary unit \(i\), in this case is going to be positive \(7\). Now you might look at this number and think, couldn't you just write that number as \(7i\)? That zero isn't really doing anything. And you're right. I could just write this as \(7i\). But we still need to know that if we're looking at this as a complex number, it still has a real part. It's just 0.

Let’s look at another example. So I have \(2+0i\) over here. So what do you think the real part of this number is? Well, since this \(2\) is out here by itself, it's not multiplying my imaginary unit. \(2\) is going to be the real part of my number, \(a\). Then looking at \(b\), so the imaginary part, the part that is multiplying my imaginary unit, in this case is just 0. So, again, you might be looking at this number thinking, isn't that just a real number? Couldn't I just write that as \(2\)? And you're right. Again, I could just write this as \(2\). But remember, if we're looking at this as a complex number, it still has an imaginary part. It's just 0. So that's all for this one, and I'll see you in the next video.

Identify the real and imaginary parts of each complex number. $-4-9i$

• A. $a = -9, b = -4$
• B. $a = -4, b = -9$
• C. $a = 4, b = 9$
• D. $a = -4, b = 9$

Identify the real and imaginary parts of each complex number. $3+2i\sqrt3$

• A. $a=3,b=2\sqrt3$
• B. $a = 3, b = 2$
• C. $a=2\sqrt3,b=3$
• D. $a=3,b=\sqrt3$

Write the complex number in standard form. $\frac{9+\sqrt{-16}}{3}$

• A. $3+4i$
• B. $9+16i$
• C. $3+\frac43i$
• D. $\frac{13i}{3}$

Adding and Subtracting Complex Numbers

Hey, everyone. Now that we know what complex numbers are, we can perform operations on them like addition and subtraction. Now I know you may be thinking that this is just yet another new skill you're going to have to learn, but adding and subtracting complex numbers is actually exactly the same as adding and subtracting algebraic expressions. So we're just going to take a skill that we already know and apply it to our complex numbers. Let's go ahead and jump in.

So just like with algebraic expressions, when you add and subtract complex numbers, you're just going to combine like terms. So when I'm given algebraic expressions to add, let's say 4 + 8x plus 2 + 3x, then I just need to remove my parentheses and then combine any like terms I have. So here I have 4 and 2, both constants that combine to 6, and then I have 8x and 3x, both x terms that combine to 11x. So when I'm working with algebraic expressions, I have some constants and some terms with variables that get combined to give me my final solution.

When I'm working with complex numbers, I still have constants, but instead of terms with a variable, I have terms with an i, my imaginary unit. So I'm really just going to treat that i as though it is a variable. So let's look at adding these complex numbers. So here I have \(4+8i+2+3i\). So let's go ahead and get rid of all of those parentheses. So I have \(4 + 8i + 2 + 3i\). And now I can just combine like terms. So I have constants \(4\) and \(2\), which are going to combine to give me \(6\). And then I have my imaginary terms, \(8i\) and \(+3i\), which are going to combine to give me \(+11i\).

Now one thing that we need to consider whenever we're adding or subtracting complex numbers is that we always want to express our final answer in standard form. Now remember, standard form for complex numbers is just \(a+bi\). So I got \(6 + 11i\) here, which is already in that standard form. So my final answer is \(6 + 11i\).

Let's take a look at subtracting some complex numbers. So, again, I have \(4 + 8i\), but now minus \(2 + 3i\). So let's go ahead and remove those parentheses. So I have \(4 + 8i\). But now since I'm subtracting, I need to make sure that I distribute that negative into my parentheses here. So I'm left with a minus \(2\) now and a minus \(3i\). Now I can simply combine like terms as I did before. So, again, my constants, \(4\) and then negative \(2\), these will combine to give me a positive \(2\). And then I have \(8i\) and negative \(3i\). So \(8i\) and negative \(3i\) are gonna combine to give me a positive \(5i\).

Now, again, we want to make sure that our answer is in standard form here, \(a + bi\), and it is. So my final answer is \(2 + 5i\). And that's all there is to adding and subtracting complex numbers. I'll see you in the next video.

Find the difference. Express your answer in standard form. $\left(2+8i\right)-\left(4-i\right)$

• A. $-2+9i$
• B. $6+7i$
• C. $2+7i$
• D. $2-9i$

Find the sum. Express your answer in standard form. $5\left(4+7i\right)+6\left(3-2i\right)$

• A. $7+5i$
• B. $38+23i$
• C. $2+47i$
• D. $7+9i$

Multiplying Complex Numbers

Hey, everyone. So not only can we add and subtract complex numbers, but we can also multiply them. Now, multiplication might seem a bit more intimidating than adding or subtracting, but we're actually still going to multiply our complex numbers the same way we multiply algebraic expressions. So, we're again going to take a skill that we already know and apply it to our complex numbers. And don't worry, I'm still going to walk you through it step by step. So let's go ahead and take a look. Now, algebraic expressions are always multiplied one of two ways, so complex numbers will be the same. We either distribute or we use foil. Now when we either distribute or foil, we're always going to end up with an I2 term. Now we know that I2 is just equal to negative one, so I'm going to use that to simplify my final answer. So let's go ahead and just jump right into an example. Looking at my first expression over here, I have 3I times 7 minus 2I. Now my very first step is going to be to either distribute or foil. And looking at my expression here, I have this 3I by itself. So since it's just one term, I'm multiplying this other complex number, I'm gonna go ahead and distribute. That seems like the best choice here. So, I'm gonna distribute this 3I into my 7 and my negative 2I in order to get 21I. And then 3I times negative 2I is gonna give me negative 6 I2. Make sure when you're multiplying an I by an I, you get I2. So step number 1 is done, and I can go ahead and move on to step 2, which is to apply the fact that I2 equals negative one. So looking at my expression here, I have 21I minus 6 I2. So I have my I2 right here, and I need to take this whole term and simplify it using I2 equals negative one. So this negative 6 I2 just becomes negative 6 times negative 1, and negative 6 times negative 1 is just positive 6. So I can bring my 21I down here, and I have 21I+6. So I've completed step 2. And step 3 is to combine my like terms. Now looking over here, I don't have any like terms that need to get combined. So step 3 is also done and this is my final answer. But I wanna make sure that I express my answer in standard form. So here at 21I+6, and I know that standard form is a+bi. So I'm gonna go ahead and flip this around in order to get 6 +21I, and this is gonna be my final solution. So let's go ahead and look at another example. So over here, I have negative 6 +2I times 3+4I. Let's go ahead and start with step 1, which is either to distribute or foil. Now since I have two complex numbers that each have two terms in them, it looks like foiling is gonna be my best option for step 1. So let's go ahead and foil. So I'm gonna start with my first term, so negative 6 times 3 is gonna give me negative 18. And then I have my outside terms, negative 6 times 4, which is gonna give me or times 4I, which is gonna give me negative 24I. Then I have my inside terms, 2I times 3 is gonna give me positive 6I. And then lastly, my last terms, which is 2I times 4I. Now this, since I'm multiplying two I terms, I'm gonna end up with an I squared. This is gonna give me plus 8 I2. Okay. So we have completed step number 1. Let's go ahead and move on to step 2, which is to apply that I2 equals negative one. Now I definitely have an I squared term over here. I have this 8I squared. So I'm gonna go ahead and simplify this whole term into 8 times negative one. Now 8 times negative one is just negative 8. So I have applied my I2 equals negative 1, and I can go ahead and pull all of my other terms down as well. So I can bring my negative 18, and then I have both of my I terms, negative 24I and positive 6I. So now step 3 is to combine all of my like terms, and I have some like terms that need to get combined here. So negative 18 and negative 8 are going to combine to give me negative 26. And then my other like terms are these I terms, negative 24 and positive 6I, which are gonna combine to give me negative 18I. So my final answer, I've completed step 3. I've combined my like terms. I have negative 26 minus 18I. And I, of course, wanna check that this is in standard form a+bi and it is. So I'm good to go and this is my final answer. That's all for this video. I'll see you in the next one.

Perform the indicated operation. Express your answer in standard form. $\left(3+8i\right)^2$

• A. $-55+48i$
• B. $9+64i$
• C. $24i$
• D. $9+24i$

Find the product. Express your answer in standard form. $2i\left(9-4i\right)\left(6+5i\right)$

• A. $8+18i$
• B. $54-20i$
• C. $54-40i$
• D. $-42+148i$

Complex Conjugates

Hey, everyone. After learning how to multiply complex numbers, you may think that dividing complex numbers is up next. And you're right, but before we learn how to divide complex numbers, we actually need to learn about something called the complex conjugate. Now, the complex conjugate might seem a little abstract or maybe even a little useless at first, but I'm going to show you exactly what the complex conjugate is and how we're going to use it in a way that will help us to divide complex numbers. So let's go ahead and get started. Now, in order to find the complex conjugate of some complex number, we simply need to reverse the sign of the imaginary part of our complex number. So if I have some complex number a+bi, I'm going to look at that imaginary part, so in this case, positive b, and I'm simply going to reverse that sign. So the complex conjugate of a+bi is a-bi. And this is going to work in the reverse direction as well. So if I wanted to find the conjugate of a-bi, I would again look at that imaginary part and simply reverse the sign. So the complex conjugate of a-bi is simply a+bi. So let’s take a look at a couple of examples and identify some complex conjugates.

So looking at my first example here, I have 1+2i. Now, I want to look at that imaginary part of my number, so in this case, positive 2, and simply reverse that sign. So the real part of my number is going to stay the same. That doesn't get changed, so I'll have a 1. And then that positive 2 is going to reverse sign into a negative 2. So the complex conjugate of 1+2i is 1-2i. Let’s look at another example. So if I have 1-2i, which is just the answer that we just got a last one, and I want to find that complex conjugate, I, again, am just going to look at that imaginary part. So here I have a negative 2, and I'm going to reverse that sign. So, again, my real part stays the same. I still have 1. And then my negative 2 is going to reverse sign into a positive 2. And of course, my I on the end there. So the complex conjugate of 1-2i is 1+2i. And you might notice that that's just the reverse of what we did in part a. Let's take a look at one more example here.

So here I have -1+2i. Now I just again want to look at that imaginary part, in this case, positive 2, and reverse that sign. Now the real part of our number is going to stay the same even if it's negative. So this negative one is going to stay a negative one. And then my positive 2, since that's my imaginary part, is going to reverse sign to negative 2. So the complex conjugate of -1+2i is -1-2i. Now what do you think will happen if I take a complex number and its conjugate and I multiply them by each other? Well, let's take a look.

So if I have a complex number, so here I have 2+3i, and I multiply it by its conjugate, 2-3i, I'm going to need to foil. So let's go ahead and do that. So, I need to take my first term, 2 times 2, and that will give me 4. And then my outside terms, 2 times negative 3 I, that’s going to give me negative 6 I. And then my inside terms, 3 I times 2 is going to give me a positive 6 I. And then, of course, my last terms, positive 3i times negative 3i is going to give me negative 9i2. Now whenever we multiply complex numbers, remember, we need to look for that I squared term. So here I have negative 9i2, and this just becomes negative 9 times negative 1, which we know negative 9 times negative 1 is just positive 9. And then I can bring down all of my other terms. So my 4 comes back down, and then I have negative 6 I, positive 6 I. So looking at this, you might notice that I have a negative 6 I and a positive 6 I in the middle there, and you might notice that these are going to cancel out. So if I have minus 6i, plus 6i, those are gone. Those are going to get canceled. So I'm just left with 4 + 9. So the like terms I need to combine, 45 are going to combine to give me 13. So 2+3i, my complex number times its conjugate, 2-3i, gave me a real number. And this is going to happen any time I multiply complex conjugates. So, multiplying complex conjugates by foiling is always going to give me a real number. Now, this is going to be really useful for us when we're dividing complex numbers, which we'll see in the next video.

Now something else that you might have noticed here is that this 4 is really just, wow, is really just the a term of my complex number squared. So this 4 is really just a2. And then my 9 that I have on the end here is just my b term, 3 squared, as well. So whenever I take complex conjugates and multiply them by each other, a+bi times a-bi, I'm really just going to get a2+b2. Now this is not something that you have to memorize but it can be a helpful shortcut if you do remember it. That's all for complex conjugates, I'll see you in the next video.

Find the product of the given complex number and its conjugate. $4-5i$

• A. $16$
• B. $41$
• C. $25$
• D. $20$

Find the product of the given complex number and its conjugate. $-7-i$

• A. $50$
• B. $14$
• C. $49$
• D. $1$

Dividing Complex Numbers

Now that we know how to add, subtract, and multiply complex numbers, let's learn how to divide them. Whenever I divide by a complex number, I'm always going to end up with a fraction that has an i in the bottom. So, I'll have something, doesn't really matter what, in the numerator divided by a complex number that has a term with i in it. Now looking at this fraction, I'm not really sure how to simplify it and come to a solution that actually makes any sense. So, this i in the denominator is actually bad. I don't know how to deal with it, and I want to get rid of it in any way that I can. So how am I going to take this complex denominator and turn it into something real? We just learned that if we have a complex number and multiply, I'm going to go ahead and walk you through how to use the complex conjugate to divide complex numbers and come to a solution. Let's go ahead and jump right into an example.

I want to find the quotient of these numbers: 31+2i. The first thing I want to do is get rid of that i in the denominator. So, my very first step is actually going to be to multiply both the top and the bottom by the complex conjugate of the bottom. Since the bottom is the number that I want to get rid of, I want to get rid of the i in that, I'm going to multiply by the complex conjugate. The complex conjugate of 1+2i is just going to take that imaginary part and flip the sign. This is going to be 1 minus 2i. Now, if I'm multiplying the bottom of my fraction by that, I need to also multiply the top of my fraction as well so that I'm not really changing the value of anything.

So let's go ahead and expand this out. With my numerator, I can take this 3 and distribute it into that complex conjugate, which will give me 3-6i. Now with the denominator, since I have two complex numbers, I can go ahead and use FOIL here. The first term is 1×1 which will give me 1. The outside term, 1×-2i, will give me a negative 2i. Then my inside term, 2i×1, gives me positive 2i. And my last term, 2i×-2i, will give me -4i2. We know that whenever we're left with an i² term, we can further simplify that. So, this -4i2 is really just negative 4 times negative 1, which we know is just positive 4.

So let's go ahead and rewrite our fraction. The numerator is still just 3-6i. And then our denominator here, let's look at what terms we have left. Well, I have this one, and then I have minus 2i plus 2i, but I know that those middle terms are just going to cancel out, so I don't have to worry about them. And then I have this plus 4. So, I'm simply just left with 1 plus 4, and 1+4 is just 5. So, my denominator here is 5.

Okay, so we have completed step number 1. We have multiplied the top and bottom by our complex conjugate and simplified as much as we can. Now let's move on to step number 2, which is actually to expand our fraction further into the real and imaginary parts. So I want to take the real part and I want to split it from my imaginary part. So looking at my fraction over here, I can just take the numerator and split it, keeping that denominator on both terms. So I will really just end up with 35 minus 65i. So I have expanded my fraction into my real and imaginary parts. Step 2 is done.

Finally, step 3, which is to simplify our fraction to our lowest terms. I have 35 minus 65i, and this is actually already in its lowest terms so that means that this is just my solution. I've completed step 3, I'm completely done, and I have my solution. That's all there is to dividing complex numbers. Let me know if you have any questions.

Find the quotient. Express your answer in standard form.  

$\frac{6+i}{4-2i}$

• A. $\frac{11}{10}+\frac45i$
• B. $\frac65+\frac45i$
• C. $\frac{11}{10}-\frac45i$
• D. $22+16i$

Find the quotient. Express your answer in standard form. 

$\frac{-5+3i}{-7-4i}$

• A. $\frac35+\frac45i$
• B. $18i$
• C. $23-41i$
• D. $\frac{23}{65}-\frac{41}{65}i$

Here’s what students ask on this topic:

What is a complex number and how is it represented?

A complex number is a number that combines a real part and an imaginary part. It is represented in the standard form $a+bi$, where $a$ is the real part and $b$ is the imaginary part. The imaginary unit $i$ is defined as $\sqrt{-1}$. For example, in the complex number $3+2i$, $3$ is the real part and $2$ is the imaginary part.

How do you add and subtract complex numbers?

Adding and subtracting complex numbers is similar to combining like terms in algebra. For addition, you add the real parts and the imaginary parts separately. For example, to add $4+8i$ and $2+3i$, you get $6+11i$. For subtraction, you subtract the real parts and the imaginary parts separately. For example, to subtract $2+3i$ from $4+8i$, you get $2+5i$.

How do you multiply complex numbers?

To multiply complex numbers, you use the distributive property or the FOIL method. For example, to multiply $-6+2i$ and $3+4i$, you apply FOIL: $-6\times 3+-6\times 4i+2i\times 3+2i\times 4i$. This results in $-18-24i+6i+8i2$. Since $i2=\; -1$, the final result is $-26-18i$.

What is the complex conjugate and how is it used in division?

The complex conjugate of a complex number $a+bi$ is $a-bi$. It is used in division to eliminate the imaginary unit from the denominator. For example, to divide $3$ by $1+2i$, multiply both the numerator and the denominator by the conjugate $1-2i$. This results in $3\times 1-3\times 2i$ over $1\times 1-2i\times 2i$, which simplifies to $3-6i$ over $5$, or $\mathrm{3/5}-\mathrm{6/5}i$.

How do you simplify the expression involving $i2$?

To simplify an expression involving $i2$, use the fact that $i2=\; -1$. For example, in the expression $3i2$, replace $i2$ with $-1$, resulting in $3\times -1=-3$. This simplification is crucial when performing operations like multiplication and division with complex numbers.