2-3i / 5-i Would I multiply both by the denominator and then FOIL?
would you like to write this in standard form?
if so multiply the bottom and the top by the bottom's conjugate
the conjugate of a+b is a-b the conjugate of a-b is a+b so what is the conjugate of 5-i?
i+5?
right
So multiple top and bottom by i+5 and then foil?
(2-3i)(5+i) is the top --------- (5-i)(5+i) is the bottom now lets multiply
2*5+2i-15i-3i^2 is the top ________________ 5^2-i^2 is the bottom but what is i^2?
-1
right! so we have 10-13i-3(-1) _____________ 25-(-1) 10-13i+3 _________ 25+1 13-13i ______ 26
1-i ____ 2
25i isn't 25(-1)? making -25.
right 25i isn't 25(-1) but 25i^2 is 25(-1)=-25 but where did you get 25i or 25i^2
Oooooh, no I get what you mean now. The denominator was 25-i^2. So the final answer was/is 13-13i ------ 26 which simplified to: 1-i/2 ?
13/26-13i/26 1/2-i/2 because 26 and 13 are both divisible by 13
Right, so 1-i --- 2 ? Sorry if I'm beating the dead horse here.
Thank you so much.
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