# Raising complex numbers to a power

• Oct 9th 2013, 07:02 PM
KevinShaughnessy
Raising complex numbers to a power
Hi,

I'm using j when I know some textbooks use i, I'm not sure what the difference is if there is any.

I have the question (4j^5 - 5j^4 + 2j^3 -3j^2)^2 and it's supposed to evaluate to -8j, I just don't remember how to get that answer. To me it seems that something like 4j^10 could be considered as 4j^2 since there are 5 pairs of -1 resulting from multiplying j by j, so 4j^10 would be 4(-1). That does not get me the right answer though. What is the flaw in my thinking?

Thanks!
• Oct 9th 2013, 08:33 PM
Prove It
Re: Raising complex numbers to a power
No there is no difference. The standard convention is to use "i" (for imaginary number), but some texts use "j" because "i" is being used for another purpose (I know this is the convention used by electrical engineers).

Anyway, to help you with your problem, recall that j^2 = -1, which means j^3 = -j, j^4 = 1, and then they repeat. That should simplify the expression greatly before trying to square.
• Oct 9th 2013, 08:39 PM
votan
Re: Raising complex numbers to a power
Quote:

Originally Posted by KevinShaughnessy
Hi,

I'm using j when I know some textbooks use i, I'm not sure what the difference is if there is any.

I have the question (4j^5 - 5j^4 + 2j^3 -3j^2)^2 and it's supposed to evaluate to -8j, I just don't remember how to get that answer. To me it seems that something like 4j^10 could be considered as 4j^2 since there are 5 pairs of -1 resulting from multiplying j by j, so 4j^10 would be 4(-1). That does not get me the right answer though. What is the flaw in my thinking?

Thanks!

Usually j is used in electricity where i is for the electric current.

Conxtruct a table to yourself like this

j^0 = j^4 = j^8 =j^(4*n) = 1 n is integer

Continue the table that starts with
j^1 =
j^2 =
j^3 =

• Oct 10th 2013, 07:47 PM
KevinShaughnessy
Re: Raising complex numbers to a power
I'm afraid I'm still stuck. When I break down (4j^5 - 5j^4 + 2j^3 -3j^2)^2, j^5 is the same as j, so 4j^5 can be rewritten as 4j. 5j^4 is the same as 1, so that can be rewritten as 5(1). 2j^3 is the same as negative j, so that can be rewritten as or -2j. j^2 is the same as negative one, so that can be rewritten as +3. That leaves me with (4j-5-2j+3)^2 which leads to 16j^2 -25 -4j^2 + 9 or -28.
• Oct 10th 2013, 08:30 PM
SlipEternal
Re: Raising complex numbers to a power
Simplify before squaring: (4j-5-2j+3)^2 = (2j-2)^2 or be more careful when multiplying out before simplifying: (4j-5-2j+3)(4j-5-2j+3) = 4j(4j-5-2j+3) - 5(4j-5-2j+3) -2j(4j-5-2j+3)+3(4j-5-2j+3). You missed a LOT of terms.
• Oct 12th 2013, 11:08 AM
KevinShaughnessy
Re: Raising complex numbers to a power
Quote:

Originally Posted by SlipEternal
Simplify before squaring: (4j-5-2j+3)^2 = (2j-2)^2 or be more careful when multiplying out before simplifying: (4j-5-2j+3)(4j-5-2j+3) = 4j(4j-5-2j+3) - 5(4j-5-2j+3) -2j(4j-5-2j+3)+3(4j-5-2j+3). You missed a LOT of terms.

Ah, I see where I went wrong. For some reason it didn't even occur to me that you needed to do that. Thanks a lot for your help!