inik, please help me, having trouble with the complex numbers solution you gave me: [(1 + i)(4 − i) − 3∗i] / (i∗(4 − i) ) = (4 + 4i − i + 1 − 3i) / (4i + 1) = 5 / (4i + 1) = 5 / (4i+1)
(4 - i +4i-(i^2) )/(4-i+4i- (i^2))=(4+4i-i+1)/(3i+5)=(3i+5)/(4i+1) this is based on what I wrote... got different answer! or made a mistake when I copied long expression or got wrong answer preciously :( what was the original problem - please remind me
\[\frac{(1+i)(4-i)-3i}{i(4-i)}\]?
Original problem: [(1+ i) / i] + [-3 / (4-i)]
\[\frac{5+3i-3i}{1+4i}\] \[\frac{5}{4+i}\]
oh.
based on satellite73...: my first answer is right! 3i & 3i are canceled out :)
satellite73.. it should be: 5/(4i+1) right?
\[\frac{1+i}{i}=1-i\] \[\frac{-3}{4+i}=- \frac{12}{17}-\frac{3}{17}i\]
So I solve them both seperately?
i guess i am confused as to which problem you are doing. if it is \[\frac{1+i}{i}-\frac{3}{4-i}\] then i get something completely different.
Completely different from inik, you mean?
i would compute each one, put in standard form and then subtract. that is all
it depends on the problem you are going. the one i wrote above is not what you have up top.
Then I'm very sorry for the mixup, because that it the problem.
I tried to write it out like it should have been.
ooooooooooooh i isee what you are doing. you are using \[\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bc}\]
maybe lnik is right.
i would just compute each one and then add or subtract real from real and complex from complex
And inik used a common denominator, right?
inink and satellite when you get done and if you have time and if you like i'm having trouble with this one question i will come back and give a link k?
if the problem is \[\frac{1+i}{i}-\frac{3}{4-i}\] then i get \[1-i-\frac{12}{17}-\frac{3}{17}i\] \[=\frac{5}{17}-\frac{20}{17}i\]
where you at?
http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e2862ba0b8b3d38d3b8e738
inik, did you get something like that?
i got what satellite got
Okay, thank you as well.
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