Thread: 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)

Hello,

The method is correct, but not the result :

i=i
i²=-1
i^3 = -i
i^4 = 1

I don't understand how the result is incorrect.

Shouldn't it be simply ?


What isn't right?

=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

as i've shown you.

So what's the result ?

So would it be (1^7)i ?

Learn these four numbers .
Now divide the exponent by 4 and take the remainder.
Thus because 31 divided by 4 leaves a remainder of 3.
Here why that works.

Originally Posted by eraser851

So would it be (1^7)i ?

The very first property of i is that i²=-1

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 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 , , or .

See here:

Imaginary Numbers