- #1

- 4

- 0

- Thread starter becka12
- Start date

- #1

- 4

- 0

- #2

- 147

- 0

- #3

- 4

- 0

- #4

- 147

- 0

- #5

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

Let y= zbecka12 said:

y

- #6

- 80

- 0

(z^2 + 2i)(z^2 - 2i)

then

(z^2 + 2i) = (z + 1 - i)(z - 1 + i)

(z^2 - 2i) = (z + 1 + i)(z - 1 - i)

multiply

(z + 1 - i)(z + 1 + i) = z^2 + 2z + 2

(z - 1 + i)(z - 1 - i) = z^2 - 2z + 2

and there you have your two real number quadratics

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 478

- Last Post

- Replies
- 4

- Views
- 941

- Last Post

- Replies
- 8

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 20

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 552

- Last Post

- Replies
- 3

- Views
- 2K