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## Integrated math 2

### Course: Integrated math 2 > Unit 5

Lesson 4: Adding & subtracting complex numbers# Adding complex numbers

Learn how to break down the process of adding complex numbers into simple steps. First, add the real parts. Then, do the same with the imaginary parts. The result? A new complex number! Learn and try some examples with Sal. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- Can someone give an example of adding complex numbers using fractions?(40 votes)
- (1/2 + 1/2 i) + (3/4 + 1/3 i) = 5/4 + 5/6 i

(.25 + i/10) + (.66 + i/8 ) = .91 + 18i/80 = 91/100 + 9i/40

(edited for clarity -- initially wrote 1 1/4 (one and one quarter) I don't like improper fractions but 5/4 is clearer. Thanks!)(63 votes)

- Okay but what if I had an equation like this: (-6 + 7i) + (6 - 7i)? Wouldn't the answer be 0 because everything deletes itself??(22 votes)
- Yes, that is correct. You are adding a number to the negative of that number, which always equals 0.(30 votes)

- Do you treat i as a variable?(7 votes)
- For many purposes,
`𝒾`

can be used just like any other variable, you just need to take special care when exponentiating it, since`𝒾² = -1`

, so unlike other variables, it can disappear when exponentiated.(21 votes)

- What is the concept of i?(4 votes)
- negative numbers cannot be put under the radical and be called a real number. Therefore, they created "i". i = the square root of -1(26 votes)

- Can someone help me with this complex number? (1+ 2i)^2(4 votes)
- Start by writing the multiplication explicitly:
`(1 + 2i) (1 + 2i)`

And multiply as normal:`1 + 2i + 2i + 4i^2`

Remember that`i^2 = -1`

`1 + 2i + 2i - 4`

Finally add real numbers and imaginary numbers, and you're done:`-3 + 4i`

(16 votes)

- how would you solve an equation in which the I's had different exponents? e.g. i3+i6+3i(3 votes)
- Recall that i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i. With this information we can separate the reals and imaginary numbers.

So for your example (i^3) + (i^6) + (3i) = (-i) + (-1) + (3i) = 2i - 1

I hope I answered your question correctly.(6 votes)

- Why don't we do what's in the parenthesis first? Why do we ignore them?(4 votes)
- We look at the example: (5+2i)+(3-7i).

We cannot add together 5 and 2i. This is like trying to add 5 and 2x, it is not mathematically logical.

So, because we are not able to add together the items in the parenthesis we move on and can add together like terms.(5 votes)

- so would i just be seen as a variable?(4 votes)
- Yes, when adding and subtracting complex numbers, "i" acts as a variable.(4 votes)

- the imaginary number (i) is defined such that i^2 = -1 what does i+i^2+i^3+....+i^23 equal(3 votes)
- I had similar question just few days ago, and after completing the exercise in Imaginary Unit Powers and studying, I FINALLY found THE ANSWER I was looking for.

To determine what happens to an imaginary number such as i when raised to a certain power.

Just follow these steps and you will be able to solve for the value of an imaginary number i raised to any power.

AND I REPEAT ANY POWER!

The following steps are as indicated below:

1st you divide its power by 4

2nd you replace its power by the remainder of the original power of i

3rd your answer will now be determined by the new power of your i [which is the remainder of (the original power)/4]

For instance:

if there is no remainder then the new power of i is zero so

i^0=1 Because (any number)^0=1

if the remainder is 1 then

i^1=i Since (any number)^1=itself

if the remainder is 2 then

i^2=-1

and finally, last but not least (and by the way this is as big as you're remainder will ever get), if the remainder is 3

i^3=-i Since (i^2)*i^1=-1*i=-i

For example: you have i^333 and you want to its value.

1st divide 333 by 4 and you get 83 remainder 1

2nd you replace its power by the remainder which is 1, i^1

3rd its new power will determine its value

since the remainder is 1

i^1=i Since (any number)^1=itself

Just follow these steps and you will be able to solve for the value of an imaginary number i raised to any power.(5 votes)

- are complex numbers closed under addition?(4 votes)
- Yes, complex numbers are closed under addition. For any two complex numbers, (a + bi) + (c + di) = (a + c) + (b + d)i which is also a complex number.(4 votes)

## Video transcript

We're asked to add the
complex number 5 plus 2i to the other complex
number 3 minus 7i. And as we'll see, when we're
adding complex numbers, you can only add the
real parts to each other and you can only add the
imaginary parts to each other. So let's add the real parts. So we have a 5 plus a 3. And then the imaginary
parts-- we have a 2i. So plus 2i. And then we have a negative
7i, or we're subtracting 7i. So minus 7i right over here. And 5 plus 3-- that's
pretty straightforward. That's just going to be 8. And then if I have
two of something and from that I subtract
seven of that something-- and in this case,
the something is the imaginary
unit, the number i. If I have two i's and I take
away seven i's, then I have negative five i's. 2 minus 7 is negative 5. So then I have negative 5i. So when you add these
two complex numbers, you get 8 minus 5i. You get another complex number. It has a real part
and an imaginary part.