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sologuitar

Member

Registered: 2022-09-19

Posts: 467

Standard Form a + bi

Write each expression in the standard form a + bi.

A. i^(23)

I say i^(23) can be expressed as (i^5)(i^4)(i^3).

i^5 = i

i^4 = 1

i^3 = -i

(i)(1)(-i) = (i)(-i) = -(i^2) = i(-1) = -I

You say?

B. i^(-15)

I think the first step is to remove the negative power.

So, i^(-15) = 1/(i^15)

i^(15) = (i^5)^3

i^5 = i

I now have 1/(i)^3 = 1/(i^3) = 1/-i

My answer = 1/-I.

Textbook answer = I

What did I do wrong with B?

Last edited by sologuitar (2023-11-21 03:47:22)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Standard Form a + bi

I say i^(23) can be expressed as (i^5)(i^4)(i^3).

add the powers  (i^5)(i^4)(i^3) = i^(5+4+3)  ≠  i^23

i^4 = (i x i) x (i x i) = -1 x -1 = 1

23 = 4 + 4 + 4 + 4 + 4 + 3

so i^23 = i^3

The second one is ok up to i^5 = i

i^3 = (i x i) x i = -i

so you have 1/(-i)  If you times top and bottom by -i you can simplify this.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

amnkb

Member

Registered: 2023-09-19

Posts: 253

Re: Standard Form a + bi

Write each expression in the standard form a + bi.

A. i^(23)

B. i^(-15)

sologuitar

Member

Registered: 2022-09-19

Posts: 467

Re: Standard Form a + bi

I say i^(23) can be expressed as (i^5)(i^4)(i^3).

add the powers  (i^5)(i^4)(i^3) = i^(5+4+3)  ≠  i^23

i^4 = (i x i) x (i x i) = -1 x -1 = 1

23 = 4 + 4 + 4 + 4 + 4 + 3

so i^23 = i^3

The second one is ok up to i^5 = i

i^3 = (i x i) x i = -i

so you have 1/(-i)  If you times top and bottom by -i you can simplify this.

Bob

I see now what I did wrong.

sologuitar

Member

Registered: 2022-09-19

Posts: 467

Re: Standard Form a + bi

Write each expression in the standard form a + bi.

A. i^(23)

B. i^(-15)

Nicely-done but try to explain what you're doing along the way.