Here's the question you clicked on:
Sinz
Simplify the following quotient of complex numbers into the form a + bi. -9+8i/1+2i
You going to mutiply the numorator and the denominatior by the conjugate. Does that help?
so multiply -9 and 1 right
multiply top and bottom of the quotient by the conjugate of 1 + 2i which is 1 - 2i
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
without changing the number
(-9 + 8i)(1-2i) ------------ (1+2i)(1-2i) the denominator simplifies to 1 + 4 = 5
so I take and combine like terms right?
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
after I get the )-9+8i)(1-2i)
so it will be -9*1 and 8i*-2i?
yes - u now simplify the top expression then divide by 5
no remember (a+b)(a+b) = a^2 + 2ab + b^2
so -9 - 16i/5 is what it should look like right
nope. your top term is off. Try using FOIL on the top again and see if you get somthing else.
okay lemme see (-9+8i)(1-2i) combine like terms -9*1=-9 8i*2i=16i total is -9+16i/5 is that right
(-9 + 8i)(1-2i) = -9 + 18i + 8i + 16 = 7 + 26i the second bracket is first multiplied by -9 then its multuplied by + 81
oh okay wasn't seeing how it was done that is why I was confuzzled
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
final result is 7/5 + (26/5) i
okay so in algebra they try to change things up so it confuses you?
Thank you Jimmy you was very helpful as was you Walleye