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## Algebra (all content)

### Course: Algebra (all content)ย >ย Unit 16

Lesson 4: Adding & subtracting complex numbers

Sal adds (5+2i) and (3-7i) . 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?
• (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!)
• Okay but what if I had an equation like this: (-6 + 7i) + (6 - 7i)? Wouldn't the answer be 0 because everything deletes itself??
• Yes, that is correct. You are adding a number to the negative of that number, which always equals 0.
• What is the concept of i?
• negative numbers cannot be put under the radical and be called a real number. Therefore, they created "i". i = the square root of -1
• Do you treat i as a variable?
• For many purposes, `๐พ` can be used just like any other variable, you just need to take special care when exponentiating it, since `๐พยฒ = -1`, so unlike other variables, it can disappear when exponentiated.
• Can someone help me with this complex number? (1+ 2i)^2
• 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`
• how would you solve an equation in which the I's had different exponents? e.g. i3+i6+3i
• 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

• Why don't we do what's in the parenthesis first? Why do we ignore them?
• 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.
• the imaginary number (i) is defined such that i^2 = -1 what does i+i^2+i^3+....+i^23 equal
• 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.
``(6 + 3๐) - (-4 -2๐) - 7๐6 + 3๐ + 4 + 2๐ - 7๐6 + 4 + 3๐ + 2๐ - 7๐(6 + 4) + (3 + 2 - 7)๐10 - 2๐``