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Mathematics 16 Online
OpenStudy (anonymous):

2+i/3-2i

OpenStudy (anonymous):

multiply with the conjugate of 3-2i??

OpenStudy (anonymous):

multiply numerator and denominator with 3+2i

OpenStudy (anonymous):

The same question with mapelletah. If it's (2+i)/(3-2i),then it can be calculated as follow: (2+i)/(3-2i)=[(2+i)*(3+2i)]/[(3+2i)*(3-2i)]=(4+7i)/13

OpenStudy (anonymous):

i multiplied it and got 6+4i+3i+2i?

OpenStudy (anonymous):

2+i(3+2i)

OpenStudy (anonymous):

nope, it should be 6+4i+3i+2i^2 6+10i-2 as i^2=-1

OpenStudy (anonymous):

how do you get 10i?

OpenStudy (anonymous):

sry typing error 7i

OpenStudy (anonymous):

6+7i^2?

OpenStudy (anonymous):

6+7i-2

OpenStudy (anonymous):

in back of the book it say (4/13)+(7/13i)

OpenStudy (anonymous):

*says

OpenStudy (anonymous):

6-2+7i is numerator denominator is ( 3-2i)*(3+2i)=9-(4i^2)=9+4=13

OpenStudy (anonymous):

which is exactly the answer just think a bit

OpenStudy (anonymous):

excuse me? i get how you got 9 from multiplying them but how did you get 4 from 9-(4i^2)?

OpenStudy (anonymous):

i^2=-1 so 4i^2=-4 we have 9-4i^2=9+4=13

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

does that help

OpenStudy (anonymous):

alot better

OpenStudy (anonymous):

great

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