# Thread: Simplifying Complex Numbers (i)

1. ## 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)

2. Hello,

The method is correct, but not the result :

i=i
i²=-1
i^3 = -i
i^4 = 1

3. I don't understand how the result is incorrect.

Shouldn't it be simply $i$?

What isn't right?

4. =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 $i^{31}=i^{28}i^3 = (i^4)^7 i^2 i$

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

So what's the result ?

5. So would it be (1^7)i ?

6. Learn these four numbers $i,\,i^2 = - 1,\,i^3 = - i,\,i^4 = 1$.
Now divide the exponent by 4 and take the remainder.
Thus $i^{31} = i^3 = - i$ because 31 divided by 4 leaves a remainder of 3.
Here why that works.
$i^{31} = i^{28 + 3} = \left( {i^4 } \right)^7 \left( {i^3 } \right) = \left( 1 \right)^7 \left( {i^3 } \right) = - i$

7. Originally Posted by eraser851
So would it be (1^7)i ?
The very first property of i is that i²=-1

8. 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 $i^4$ 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 $i^1$, $i^2$, or $i^3$.

See here:

Imaginary Numbers