Question 1

Can someone give an example of adding complex numbers using fractions?

Accepted Answer

(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!)

Question 2

Okay but what if I had an equation like this: (-6 + 7i) + (6 - 7i)? Wouldn't the answer be 0 because everything deletes itself??

Accepted Answer

Yes, that is correct. You are adding a number to the negative of that number, which always equals 0.

Question 3

Do you treat i as a variable?

Accepted Answer

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.

Question 4

What is the concept of i?

Accepted Answer

negative numbers cannot be put under the radical and  be called a real number. Therefore, they created "i". i = the square root of -1

Question 5

Can someone help me with this complex number? (1+ 2i)^2

Accepted Answer

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`

Question 6

Why don't we do what's in the parenthesis first?  Why do we ignore them?

Accepted Answer

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.

Question 7

are complex numbers closed under addition?

Accepted Answer

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.

Question 8

what if it has an i outside (6+3i)-(-4-2i)-7i?

Accepted Answer

Then you are adding 3 complex numbers. So if you want you can add the first 2, and then add the last one, or you can do it all at the same time, adding all the real and all the imaginary parts:
```
(6 + 3𝑖) - (-4 -2𝑖) - 7𝑖
6 + 3𝑖 + 4 + 2𝑖 - 7𝑖
6 + 4 + 3𝑖 + 2𝑖 - 7𝑖
(6 + 4) + (3 + 2 - 7)𝑖
10 - 2𝑖
```

Question 9

how would you solve an equation in which the I's had different exponents? e.g. i3+i6+3i

Accepted Answer

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.

Question 10

the imaginary number (i) is defined such that i^2 = -1 what does i+i^2+i^3+....+i^23 equal

Accepted Answer

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.

Algebra (all content)

Course: Algebra (all content) > Unit 16

Adding complex numbers

Want to join the conversation?

Video transcript