# I need help with complex numbers and real and imaginary.

• Dec 9th 2009, 03:52 PM
julienumbers
I need help with complex numbers and real and imaginary.
I'm in Acc Math 1, which is basically Honors Algebra 1. I have a quiz tomorrow and I really don't understand what we're learning because my teacher isn't very good.

(3+5i)/2i

(-1+7i)/(3-2i)

(8+3i)/(3-i)

Find real numbers x and y to make each of the following statements true.
4x-4yi = 1-8i
x=1/4 and y=2

Finally, I know you have to FOIL this problem, but I still can't seem to get a logical answer.
(3-8i)(-6-5i)

I know there are a lot of problems, but even if someone helps me on just one I'll be very grateful. I'm lost, and I feel like I'm the only one in the class who is. Thanks so much! :)
• Dec 9th 2009, 04:03 PM
Prove It
Quote:

Originally Posted by julienumbers
I'm in Acc Math 1, which is basically Honors Algebra 1. I have a quiz tomorrow and I really don't understand what we're learning because my teacher isn't very good.

(3+5i)/2i

(-1+7i)/(3-2i)

(8+3i)/(3-i)

Find real numbers x and y to make each of the following statements true.
4x-4yi = 1-8i
x=1/4 and y=2

Finally, I know you have to FOIL this problem, but I still can't seem to get a logical answer.
(3-8i)(-6-5i)

I know there are a lot of problems, but even if someone helps me on just one I'll be very grateful. I'm lost, and I feel like I'm the only one in the class who is. Thanks so much! :)

Are you asking to simplify these?

1. $\frac{3 + 5i}{2i} = \frac{i(3 + 5i)}{2i^2}$

$= \frac{3i + 5i^2}{-2}$

$= \frac{-5 + 3i}{-2}$

$= \frac{5 - 3i}{2}$.

2. $\frac{-1 + 7i}{3 - 2i} = \frac{(-1 + 7i)(3 + 2i)}{(3 - 2i)(3 + 2i)}$

$= \frac{-3 - 2i + 21i + 14i^2}{9 + 6i - 6i - 4i^2}$

$= \frac{-3 + 19i - 14}{9 + 4}$

$= \frac{-17 + 19i}{13}$.

3. $\frac{8 + 3i}{3 - i} = \frac{(8 + 3i)(3 + i)}{(3 - i)(3 + i)}$

$= \frac{24 + 8i + 9i + 3i^2}{9 + 3i - 3i - i^2}$

$= \frac{24 + 17i - 3}{9 + 1}$

$= \frac{21 + 17i}{10}$.

4. $4x - 4iy = 1 - 8i$

Equating real and imaginary components gives

$4x = 1$ and $-4y = -8$.

Therefore $x = \frac{1}{4}$ and $y = 2$.

5. $(3-8i)(-6-5i) = -18 -15i + 48i + 40i^2$

$= -18 + 33i - 40$

$= -58 + 33i$.
• Dec 9th 2009, 04:14 PM
julienumbers
Quote:

Originally Posted by Prove It
Are you asking to simplify these?

1. $\frac{3 + 5i}{2i} = \frac{i(3 + 5i)}{2i^2}$

$= \frac{3i + 5i^2}{-2}$

$= \frac{-5 + 3i}{-2}$

$= \frac{5 - 3i}{2}$.

2. $\frac{-1 + 7i}{3 - 2i} = \frac{(-1 + 7i)(3 + 2i)}{(3 - 2i)(3 + 2i)}$

$= \frac{-3 - 2i + 21i + 14i^2}{9 + 6i - 6i - 4i^2}$

$= \frac{-3 + 19i - 14}{9 + 4}$

$= \frac{-17 + 19i}{13}$.

3. $\frac{8 + 3i}{3 - i} = \frac{(8 + 3i)(3 + i)}{(3 - i)(3 + i)}$

$= \frac{24 + 8i + 9i + 3i^2}{9 + 3i - 3i - i^2}$

$= \frac{24 + 17i - 3}{9 + 1}$

$= \frac{21 + 17i}{10}$.

4. $4x - 4iy = 1 - 8i$

Equating real and imaginary components gives

$4x = 1$ and $-4y = -8$.

Therefore $x = \frac{1}{4}$ and $y = 2$.

5. $(3-8i)(-6-5i) = -18 -15i + 48i + 40i^2$

$= -18 + 33i - 40$

$= -58 + 33i$.

Oh, my problem was the whole conjugate part. Thank you so much! I get it now. I just needed some help. :)