Question

Simplify the following quotient of complex numbers into the form a + bi. -9+8i/1+2i

Answer 1

You going to mutiply the numorator and the denominatior by the conjugate. Does that help?

Answer 2

so multiply -9 and 1 right

Answer 3

multiply top and bottom of the quotient by the conjugate of 1 + 2i which is 1 - 2i

Answer 4

huh I am confused

Answer 5

Your goal here is to "kill" the bottom complex number. you can do this by

\[(9+8i/1+2i) * (1 - 2i / 1 - 2i)\]

remember that 1-2i / 1-2i = 1 and you can multiply anything by 1

Answer 6

without changing the number

Answer 7

okay

Answer 8

(-9 + 8i)(1-2i)
------------
(1+2i)(1-2i)

the denominator simplifies to 1 + 4 = 5

Answer 9

so I take and combine like terms right?

Answer 10

yep. you'll find that if you multiply a complex number by its conjugate then you kill all of the i terms and just be left with real numbers

Answer 11

after I get the )-9+8i)(1-2i)

Answer 12

so it will be -9*1 and 8i*-2i?

Answer 13

yes - u now simplify the top expression
then divide by 5

Answer 14

no
remember
(a+b)(a+b) = a^2 + 2ab + b^2

Answer 15

so -9 - 16i/5 is what it should look like right

Answer 16

nope. your top term is off. Try using FOIL on the top again and see if you get somthing else.

Answer 17

okay lemme see (-9+8i)(1-2i) combine like terms -9*1=-9 8i*2i=16i total is -9+16i/5 is that right

Answer 18

(-9 + 8i)(1-2i) = -9 + 18i + 8i + 16 = 7 + 26i

the second bracket is first multiplied by -9 then its multuplied by + 81

Answer 19

by +8i

Answer 20

oh okay wasn't seeing how it was done that is why I was confuzzled

Answer 21

I dont know if this will help but treat i just like you would x when your distributing! I find that thinking about it that way very helpfful

Answer 22

final result is

7/5 + (26/5) i

Answer 23

okay so in algebra they try to change things up so it confuses you?

Answer 24

Thank you Jimmy you was very helpful as was you Walleye

Answer 25

yw!

Answer 26

thats ok