Multiplying complex numbers
Sal multiplies (1-3i) by (2+5i). Created by Sal Khan and Monterey Institute for Technology and Education.
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- How would you simplify (2i)(5+4i) ??(27 votes)
i squared (i^2)= -1
So when you distribute you get 10i+ 8i^2
Since you know that i^=1, you can take 8(-1)= -8
So your final answer is: -8+10i(3 votes)
- I was wondering, Since addition and subtraction of complex numbers use traditional methods, why do complex number's require Binomial folding? not that I dislike folding it just seems a bit arbitrary at first.(15 votes)
- Complex numbers use binomial methods of multiplication because unlike real numbers, imaginary numbers have two components. Imaginary numbers are generally defined using the form a + bi where a and b are both real numbers. Due to the fact that imaginary numbers have two parts (although a can be 0) we must multiply them using by either the distributive property or by FOILing. Real numbers can also be defined by using the imaginary system of a + bi by merely having b equal to 0.(22 votes)
- What is Argand Plane in Complex Numbers? Is Khan covering it in any of the upcoming topics under complex numbers?(14 votes)
- Thank you for your question.
I copied the following from the wikipedia article on complex numbers. I had never heard the term Argand Plane prior to seeing your question.
"The complex plane is sometimes called the Argand plane because it is used in Argand diagrams. These are named after Jean-Robert Argand (1768–1822), although they were first described by Danish land surveyor and mathematician Caspar Wessel (1745–1818). Argand diagrams are frequently used to plot the positions of the poles and zeroes of a function in the complex plane."
Yes you will see plots of complex numbers in the complex plane here in Khan academy, but I'm not sure that you will ever hear that term used.(22 votes)
- I don't get how we can say that i² = -1 ?
We don't know the value of i, if we did we would have solved the calculus much faster. Where does that come from ?(4 votes)
- Hey! Well, actually in complex numbers i^2 is defined by rule as -1 and that is where we get the imaginary value of i. If you want to go the other way around, then you can simply square i although its value is imaginary: sqrt(-1)^2= -1.
In the first video on "The imaginary unit i" Sal talked about i and powers of i so you might want to check that out.
Hope this helped...(10 votes)
- what is a complex number? describe in detail(6 votes)
- A complex number is a number that can be written (that doesn't mean it is currently being expressed this way) as a + bi, where a and b are real numbers.
At least in terms of numbers you will encounter at this level of study, this means that ALL numbers are complex numbers because all numbers can be expressed in the form of a + bi.
A real number is either a rational or irrational number. All real numbers can be expressed in terms of a + 0i. Thus, a real number is a special kind of complex number, specifically one in which the coefficient (b) of i is 0 when written in a+bi form.
A nonreal complex number (often just called a nonreal number) is a complex number which, when written in the form of a + bi, the b is NOT 0. As the name implies, all numbers that are not real numbers are nonreal complex numbers.
An imaginary number is a nonreal number which, when written in the a + bi form, the a IS 0. In other words, an imaginary number can be written as 0 + bi. This means that an imaginary number is a number that can be expressed is i times some real number other than 0.
NOTE: far too many teachers and textbooks get this wrong and call all nonreal numbers "imaginary numbers", so you need to check with your teacher to make sure you use the terms the way the teacher wants, even if it is wrong.
So, in summary, real numbers and imaginary numbers are both specical kinds of complex numbers:
0 + biis the form of an imaginary number
a + 0iis the form of a real number.(12 votes)
- what if you have something like (1-i)^3?(3 votes)
- Expanding it we get
(1 - i) x (1 - i) x (1 - i)
The first two parentheses give us:
(1 -2i +i²) x (1 - i)
which simplifies . . .
= (1 -2i -1) x (1 - i)
=-2i x (1 - i)
Multiplying out gives us
which simplifies. . .
-2i + 2
= 2 - 2i
We can do a quick magnitude test to see if things are good:
To calculate the magnitude, take the root of the sum of the imaginary part squared and the real part squared. So for the original term (without the exponent):
(1 - i), the magnitude is square root( 1² + 1² ) = 2^(1/2)
The result was (2 - 2i), which has a magnitude of:
square root of (2² + 2²) = 8^(1/2) = 2^(3/2)
2^(3/2) is 2^(1/2)^3, so our answer (could) be correct.(8 votes)
- Can someone please explain the whole i^2 is equal to -1 thing? Also, why would -15^-1 be 15? Wouldn't be -1/15?(3 votes)
- (-15)⁻¹ = 1/(-15) = -1/15
𝑖 is the imaginary unit. It opens up a whole new number system called the imaginary numbers. Try to think of a number that when multiplied by itself, gives a negative number. Well a positive number times itself is a positive number. So is a negative number times itself! So it may seem that no number can satisfy this property! In fact no real number (a number in the real number system which is what you are used to working with), satisfies this property - only imaginary numbers do. This is because, we define:
𝑖 = √(-1)
𝑖² = -1
All imaginary numbers, when squared, give a negative real number. All imaginary numbers are multiples of 𝑖. The imaginary unit is analogous to 1 in the real number system - for all imaginary numbers are multiples of 𝑖. For example, 2𝑖, -44𝑖, π𝑖, or -𝑖√3. It may be hard to wrap your head around these numbers but they are just another way of thinking - just like when you were introduced to negative numbers (after all, what number can be less than nothing at all?). Comment if you have questions.(6 votes)
- Though my question sounds stupid, I want to know if we can use i as a variable.(2 votes)
- i can be treated as a variable in the usual way, in that you can freely add, subtract, multiply, divide, take roots, and exponentiate with it. It's slightly nicer than a general variable, in fact, since we can freely divide by i (since we know i≠0) and we have the simplifying property that i²=-1.
If you're asking whether you can use the symbol i to stand in for an unknown, then you can do so. But to avoid confusion, I recommend against it if you're also going to be working with the imaginary unit i.(7 votes)
- how do you know that i squared is -1?(2 votes)
- i = √(-1)
i^2 = √(-1) * √(-1) = √[(-1)(-1)] = -1
because there are two of the same factors (-1) within the radical.(5 votes)
- When he rewrote the first equation where did he get the + before the 5i from?(3 votes)
- It's from the distributive property, as Sal explained in the video:
The original expression: (1 - 3i)(2 + 5i)
The distributive property: a(b + c) = ab + ac
In this case, a = (1 - 3i), b = 2, and c = 5i
Sal distributed (1 - 3i) to the 2 and the 5i, making the expression
2(1 - 3i) + 5i(1 - 3i).(3 votes)
We're asked to multiply the complex number 1 minus 3i times the complex number 2 plus 5i. And the general idea here is you can multiply these complex numbers like you would have multiplied any traditional binomial. You just have to remember that this isn't a variable. This is the imaginary unit i, or it's just i. But we could do that in two ways. We could just do the distributive property twice, which I like a little bit more, just because you're doing it from a fundamental principle. It's nothing new. Or you could use FOIL, which you also used when you first multiplied binomials. And I'll do it both ways. So this is just a number, 1 minus 3i. And so we can distribute it over the two numbers inside of this expression. So when we're multiplying it times this entire expression, we can multiply 1 minus 3i times 2 and 1 minus 3i times 5i. So let's do that. So this can be rewritten as 1 minus 3i times 2-- I'll write the 2 out front-- plus 1 minus 3i times 5i. All I did is the distributive property here. All I said is, look, if I have a times b plus c, this is the same thing as ab plus ac. I just distributed the a on the b or the c. I distributed the 1 minus 3i on the 2 and the 5i. And then I can do it again. I have a 2 now times 1 minus 3i. I can distribute it. 2 times 1 is 2. 2 times negative 3i is negative 6i. And over here, I'll do it again. 5i times 1-- so it's plus. 5i times 1 is 5i. And then 5i times negative 3i-- so let's be careful here-- 5 times negative 3 is negative 15. And then I have an i times an i. Let me do this over here. 5i times negative 3i-- this is the same thing as 5 times negative 3 times i times i. So the 5 times negative 3 is negative 15. And then we have i times i, which is i squared. Now, we know what i squared is. By definition, i squared is negative 1. i squared, by definition, is negative 1. So you have negative 15 times negative 1. Well that's the same thing as positive 15. So this can be rewritten as 2 minus 6i plus 5i. Negative 15 times negative 1 is positive 15. Now we can add the real parts. We have a 2, and we have a positive 15. So 2 plus 15. And we can add the imaginary parts. We have a negative 6. So we have a negative 6, or a negative 6i, I should say. And then we have plus 5i. And 2 plus 15 is 17. And if I have negative six of something plus five of that something, what do I have? Or if I have five of that something and I take six of that something away, then I have negative one of that something. Negative 6i plus 5i is negative 1i, or I could just say minus i. So in this way, I just multiplied these two expressions or these two complex numbers, really. I multiplied them just using the distributive property twice. You could also do it using FOIL. And I'll do that right now really fast. It is a little bit faster. But it's a little bit mechanical. So you might forget why you're doing it in the first place. But at the end of the day, you are doing the same thing here. You're essentially multiplying every term of this first number or every part of this first number times every part of the second number. And FOIL just makes sure that we're doing it. And let me just write FOIL out here, which I'm not a huge fan of, but I'll do it just in case that's the way you're learning it. So FOIL says, let's multiply the first numbers. So that's going to be the 1 times the 2. That is the F in FOIL. Then it says, let's multiply the outer numbers times each other. So that's 1 times 5i, so plus 1 times 5i. This is the O in FOIL, the outer numbers. Then we do the inner numbers, negative 3i times 2. So this is negative 3i times 2. Those are the inner numbers. And then we do the last numbers, negative 3i times 5i. These are the last numbers. So that's all that FOIL is telling us. It's just making sure we're multiplying every part of this number times every part of that number. And then when we simplify it, 1 times 2 is 2. 1 times 5i is 5i. Negative 3i times 2 is negative 6i. And negative 3i times 5i-- well, we already figured out what that was. Negative 3i times 5i turns out to be 15. Negative 3 times 5 is negative 15. But i times i is negative 1. Negative 15 times negative 1 is positive 15. Add the real parts, 2 plus 15. You get 17. Add the imaginary parts. You have 5i minus 6i. You get negative i. And once again, you get the exact same answer.