Can't figure this out. 3-i / 8-9i

question here 3-i / 8-9i
okay so i did the conjugate. i know how to do the addition ones. but when i use the conjugate and then get the answers on top and on bottom, i come out with a really weird answer. i don't think i'm combining the right terms or something.

3-i / 8-9i

Mutiply all by 8 + 9i

so 3-i / 8-9i * 8 + 9i/8 + 9i

= (3 - i )(8 + 9i) / (8 - 9i)(8 + 9i)

= 24 - 8i + 27i -9i^2 / 64 - 81i^2


you have to realise that i^2 = -1

so

= 24 + 19i + 9 / 64 + 81 = 33 + 19i / 145


33 + 19i all divided by 145 as your answer

(3 - i)/(8 - 9i). This not what you wrote but I think it is what you meant, so here it is in excruciating detail


(3 - i)(8 + 9i)/(8 - 9i)(8 + 9i) = (24 - 8i + 27i - 9i^2)/(64 - 81i^2) =

(24 +19i +9)/(64 + 81) = (33 + 19i)/(145)

NOTE: The forward slash "/" only divides the number or expression immediately before it by the one immediately after it. Since this is a division it must be executed before any additions or subtractions in the same overall expression. That';s why we use parentheses - because what is inside them is executed before division or multiplication.

you need to multiply by (8-9i)/(8-9i)
this is the same as multiply by 1 so does not change the result however the "new" divider would be (8-9i)*(8-9i) which will be a real number (the imaginary part will be eliminated because it will have i^2).

(3-i) / (8-9i)
(3-i)(8+9i) / (8-9i)(8+9i)
(24+27i-8i-9i^2) / (64+72i-72i-81i^2)
(24+19i-9i^2) / (64-81i^2)
[24+19i-9(√-1)^2] / [64-81(√-1)^2]
[24+19i-9(-1)] / [64-81(-1)]
(24+19i+9) / (64+81)
(33+19i) / 145

excuse but i have great clean, it soft , smooth and flawless no acne or anything, i consider myself pretty thats all that matters!