SOLUTION: ((z^3)-(z^2)-4)/(z+i) i need to work it out then state the remainder. It's revision and i cant remember what to do it would be great if i could get some assistance.

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: ((z^3)-(z^2)-4)/(z+i) i need to work it out then state the remainder. It's revision and i cant remember what to do it would be great if i could get some assistance.       Log On


   

Question 288565: ((z^3)-(z^2)-4)/(z+i) i need to work it out then state the remainder.
It's revision and i cant remember what to do it would be great if i could get some assistance.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
The division process is shown below in detail.

Original expression is:

%28z%5E3-z%5E2-4%29%2F%28z%2Bi%29

First you divide z into z%5E3 to get z%5E2.
Then you multiply z%2Bi by z%5E2 to get z%5E3+%2B+z%5E2i.
Then you subtract z%5E3+%2B+z%5E2i from z%5E3+-+z%5E2 to get -z%5E2+-+z%5E2i.
Then you divide z into -z%5E2 to get -z.
Then you multiply z%2Bi by -z to get -z%5E2+-+zi.
Then you subtract -z%5E2+-+zi from -z%5E2+-+z%5E2i to get -z%5E2i+%2B+zi.
Then you divide z into -z%5E2i to get -zi.
Then you multiply z%2Bi by -zi to get -z%5E2i+-+zi%5E2.
Then you subtract -z%5E2i+-+zi%5E2 from -z%5E2i+%2B+zi to get zi+%2B+zi%5E2.
Then you divide z into zi to get i.
Then you multiply z%2Bi by i to get zi+%2B+i%5E2.
Then you subtract zi+%2B+i%5E2 from zi+%2B+zi%5E2 to get zi%5E2+-+i%5E2.
Then you divide z into zi%5E2 to get i%5E2.
Then you multiply z%2Bi by i%5E2 to get zi%5E2+%2B+i%5E3.
Then you subtract zi%5E2+%2B+i%5E3 from zi%5E2+-+i%5E2 to get -i%5E2+-+i%5E3.
Then you bring down the -4 to get -i%5E2+-+i%5E3+-+4.

Since i%5E2 = -1, you can substitute in this last expression to get:

-%28-1%29+-+%28-1%2Ai%29+-+4 which becomes:

1+%2B+i+-+4 which becomes:

i+-+3.

That's your remainder.

Your answer is:

%28z%5E3-z%5E2-4%29%2F%28z%2Bi%29 = z%5E2+-+z+-+zi+%2B+i+%2B+i%5E2 with a remainder of i+-+3.

To prove that this is correct, you need to multiply the answer by z%2Bi and then add the remainder back in to see if you can duplicate the original expression.

You do that in the following manner.

z%5E2+-+z+-+zi+%2B+i+%2B+i%5E2 * z%2Bi equals:

z%5E2+-+z+-+zi+%2B+i+%2B+i%5E2 * z plus:
z%5E2+-+z+-+zi+%2B+i+%2B+i%5E2 * i.

First we multiply z%5E2+-+z+-+zi+%2B+i+%2B+i%5E2 * z.

That becomes:

z%5E3+-+z%5E2+-+z%5E2i+%2B+zi+%2B+zi%5E2.

Then we multiply z%5E2+-+z+-+zi+%2B+i+%2B+i%5E2 * i.

That becomes:

z%5E2i+-+zi+-+zi%5E2+%2B+i%5E2+%2B+i%5E3.

Then we add:

z%5E3+-+z%5E2+-+z%5E2i+%2B+zi+%2B+zi%5E2 and z%5E2i+-+zi+-+zi%5E2+%2B+i%5E2+%2B+i%5E3 together to get:

z%5E3+-+z%5E2+-+z%5E2i+%2B+zi+%2B+zi%5E2+%2B+z%5E2i+-+zi+-+zi%5E2+%2B+i%5E2+%2B+i%5E3.

Then we combine like terms to get:

z%5E3+-+z%5E2+%2B+i%5E2+%2B+i%5E3.

z%5E2i and -z%5E2i canceled out.
zi and -zi canceled out.
zi%5E2 and -zi%5E2 canceled out.

You are left with:

z%5E3+-+z%5E2+%2B+i%5E2+%2B+i%5E3.

Since i%5E2 = -1, you can substitute in this expression to get:

z%5E3+-+z%5E2+%2B+i%5E2+%2B+i%5E3 = z%5E3+-+z%5E2+-+1+-+i.

You now need to add the remainder of i+-+3 back in.

You get:

z%5E3+-+z%5E2+-+1+-+i plus i+-+3 equals:

z%5E3+-+z%5E2+-+1+-+i+%2B+i+-+3.

Combine like results to get:

z%5E3+-+z%5E2+-+4.

The -i and the i canceled out.

Since this is the same as the original expression you started with, your division is confirmed as being successfully concluded.