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anonymous
 one year ago
Find the quotient. Write your answer in standard form.
3 + i / 3  i
A. 1
B. 1  i
C. 3/5 + 4/5 i
D. 4/5 + 3/5 i
anonymous
 one year ago
Find the quotient. Write your answer in standard form. 3 + i / 3  i A. 1 B. 1  i C. 3/5 + 4/5 i D. 4/5 + 3/5 i

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Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0multiply top and bottom both by the conjugate of the denominator (3i)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0do you know the definition of conjugate ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0conjugate example just change the sign of imaginary number a+bi conjugate is abi bi imaginary a= real so 3i conjugate is what ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it \[\frac{3+i}{3i}\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if so, multiply top and bottom by the conjugate of the denominator the conjugate of \(a+bi\) is \(abi\) and this works because \[(a+bi)(abi)=a^2+b^2\] a real number

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0first step is \[\frac{3+i}{31}\times \frac{3+i}{3+1}\] which gives you \[\frac{(3+i)(3+i)}{3^2+1^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the denominator is evidently \(10\) and the numerator is whatever you get when you multiply that out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you good from there?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry typo there i meant "first step is \[\frac{3+i}{3i}\times \frac{3+i}{3+i}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got (9 + 6i + i^2) / (9  i) I don't know where to go from there? @satellite73

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok lets back up, your denominator is wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is not \(9i\) but rather \(9+1=10\) \[3i\] has real part \(3\) and imaginary part \(1\) when you multiply \[(3+i)(3i)\] you get \[3^2+(1)^2\] or \(10\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can write it that way if you want, but don't forget \(i^2=1\) so \(9i^2=9+1=10\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So is i by itself 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you should really just be thinking that \[(a+bi)(abi)=a^2+b^2\] just like in pythagoras

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no \(i\neq 1\) but \(i^2=1\) and so \(i^2=+1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0probably the confusing part is how to interpret \(a+bi\) when you have \(3i\) in that case \(a=3,b=1\) of course when you square, you can ignore the minus sign

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0your numerator of \[9+6i+i^2\]is correct, but not complete since \(i^2=1\) then \[9+6i+i^2=9+6i1=8+6i\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0giving you an almost final answer of \[\frac{8+6i}{10}\] now cancel the common factor of 2 top and bottom

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So it's D!! I UNDERSTAND!! Thank you so much. (:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes it is D, and you are welcome
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