Simplifying Complex Numbers (i)

• Mar 18th 2008, 01:28 PM
eraser851
Simplifying Complex Numbers (i)
Okay, I think I did this right, but I'm going to post it here just to make sure.

Simplify the complex number i^31 as much as possible.

Here's what I did:

i^31
=i^30+1
=i^6*5+1
=i^1
=i

So would my final answer just be i? (assuming I did that right)
• Mar 18th 2008, 01:40 PM
Moo
Hello,

The method is correct, but not the result :

i=i
iČ=-1
i^3 = -i
i^4 = 1

;)
• Mar 18th 2008, 01:43 PM
eraser851
I don't understand how the result is incorrect.

Shouldn't it be simply $\displaystyle i$?

What isn't right?
• Mar 18th 2008, 01:53 PM
Moo
Quote:

=i^6*5+1
=i^1
This step is strange.

This means you suppose i^(6*5) = 1, which is false.

You may write that $\displaystyle i^{31}=i^{28}i^3 = (i^4)^7 i^2 i$

$\displaystyle i^4 = 1$ as i've shown you.

So what's the result ?
• Mar 18th 2008, 01:56 PM
eraser851
So would it be (1^7)i ?
• Mar 18th 2008, 01:57 PM
Plato
Learn these four numbers $\displaystyle i,\,i^2 = - 1,\,i^3 = - i,\,i^4 = 1$.
Now divide the exponent by 4 and take the remainder.
Thus $\displaystyle i^{31} = i^3 = - i$ because 31 divided by 4 leaves a remainder of 3.
Here why that works.
$\displaystyle i^{31} = i^{28 + 3} = \left( {i^4 } \right)^7 \left( {i^3 } \right) = \left( 1 \right)^7 \left( {i^3 } \right) = - i$
• Mar 18th 2008, 02:05 PM
Moo
Quote:

Originally Posted by eraser851
So would it be (1^7)i ?

The very first property of i is that iČ=-1 ;)
• Mar 18th 2008, 09:17 PM
mathceleb
eraser,

Plato is right. When dealing with Imaginary numbers, learn those 4 powers. If the power you are to raise i to exceeds 4, you want to use $\displaystyle i^4$ and see what the remainder of powers is after you factor out a 4. Either 4 divides in evenly or it doesn't. If it does not, you are left with either $\displaystyle i^1$, $\displaystyle i^2$, or $\displaystyle i^3$.

See here:

Imaginary Numbers