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Algebra 2
Course: Algebra 2 > Unit 2
Lesson 4: Adding and subtracting complex numbersSubtracting complex numbers
Learn how to break down the process of subtracting complex numbers into simple steps. Distribute the negative sign to get rid of parentheses, then add up the real and imaginary parts separately. Voila! You've got your answer in the form of a + bi. Created by Sal Khan and Monterey Institute for Technology and Education.
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I still don't know which part is imaginary and which part is real. Can someone explain it to me?(18 votes)
in a complex number a + bi, a is the real part, and bi is the imaginary part, with b not equal to 0.
e.g. 5 + 2i, 5 is real, 2i is imaginary.
6 - 3i, 6 is real, 3i is imaginary. (or -3i, if you like.)
hope this helps!(43 votes)
Why imaginary number is called imaginary number................(1 vote)
"Those numbers were once poorly understood and regarded by some as fictitious.The use of imaginary numbers was not widely accepted by many .It was left to Euler and others to use them. Euler, in his wrote in his Introduction to Algebra
Because all conceivable numbers are either greater than zero or less than 0 or equal to 0, then it is clear that the square roots of negative numbers cannot be included among the possible numbers [real numbers]. Consequently we must say that these are impossible numbers. And this circumstance leads us to the concept of such number, which by their nature are impossible, and ordinarily are called imaginary or fancied numbers, because they exist only in imagination.
In short, it is a matter of acceptance of or the lack of it which gave them the name."(14 votes)
Imaginary numbers are so simple yet complex(5 votes)
How would one solve a problem like:
(-4-√ 75)/40
or one like:
5/(7-i)
Is there a video for these types of problems?(2 votes)
The first one is almost simplified.
https://www.khanacademy.org/math/algebra/rational-exponents-and-radicals/alg1-simplify-square-roots/v/simplifying-square-roots-1
The second one is solved by rationalizing the denominator.
https://www.khanacademy.org/math/precalculus/imaginary-and-complex-numbers/complex-conjugates-and-dividing-complex-numbers/v/dividing-complex-numbers(5 votes)
So in adding, subtracting, and multiplying the imaginary unit i, we are literally just treating it like a variable?(1 vote)
For the most part. The exception is multiplication.
-- If you have a problem that requires that you multiply i*i, you need to change it to -1.
-- If you have "i" with an exponent, you can do the multiplication.
Hope this helps.(6 votes)
how do you do the problem (1+5)-7-(8i)(2 votes)- 1+5=6
6-7=-1
-(8i)=-8i
so -1-8i(3 votes)
The result of 16-18 is -2. This is calculated by subtracting 18 from 16. Is there anything else you would like to know?(4 votes)
- How do you distribute the minus sign when subtracting complex numbers?(2 votes)
Hello Sam, you can treat a complex number like any other pair of numbers. Example: -(6+18i) becomes, -6-18i. Similarly, -(-5-24i) becomes 5+24i. An easy way to think about it is to sperate the numbers, -(6+18i) is the same as: -(6)+ -(18i). For pairs it is the same. (24+5i)-(15-12i) can be re-written as: (24+5i)+(-15+12i). Hope this helps!(2 votes)
- It is something that under no circumstances can I get it. If i^2=-1 then shouldn't i be + or - sqareroot of -1? But, Sal says that it equals only the squareroot of -1, ignoring the negative version.(2 votes)
By definition i is the square root of -1. Part of this is just what mathematicians have construed and have given it a definition to mean something. It may seem odd a little, but this definition is what allows us to work with negative roots and complex numbers.(2 votes)
Is this really this simple?( I mean are there more sophisticated problems?)(0 votes)
I think the aim here is to make you understand the concept..
if you want more 'sophisticated' problems, you should try buying a book or searching on the web for them(7 votes)
Video transcript
We're asked to subtract. And we have the complex
number 2 minus 3i. And from that, we are
subtracting 6 minus 18i. So the first thing
I'd like to do here is to just get rid
of these parentheses. So we just have a bunch of
real parts and imaginary parts that we can then
add up together. So we have 2 minus 3i. And then we're subtracting
this entire quantity. And to get rid of
the parentheses, we can just distribute
the negative sign. Or another way to
think about it, we can say that this is
negative 1 times all of this. So we can just distribute
the negative sign. And negative 1 times
6 is negative 6. Let me do these in magenta. So this is negative 6. And then negative 1 times
negative 18i-- well, that's just going to be positive 18i. Negative times a
negative is a positive. And now we want to
add the real parts, and we want to add
the imaginary parts. So here's a real part here, 2. And then we have a minus 6. So we have 2 minus 6. And we want to add
the imaginary parts. We have a negative. Let me do that in
a different color. We have a negative
3i right over here. So negative 3i, or minus
3i right over there. And then we have a plus
18i or positive 18i. If you add the real parts,
2 minus 6 is negative 4. And you add the imaginary parts. If I have negative
3 of something and to that I add
18 of something, well, that's just going to leave
me with 15 of that something. Or another way you
could think about it, if I have 18 of something and
I subtract 3 of that something, I'll have 15 of that something. And in this case, the something
is i, is the imaginary unit. So this is going to be plus 15i. And we are done.