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Mathematics 14 Online
OpenStudy (lisa123):

write in the form a+bi (3-i)/(2-i) X (3+i)/(4+i)

OpenStudy (solomonzelman):

do you know what "conjugate" is?

OpenStudy (lisa123):

lol I wa about to tag you

OpenStudy (solomonzelman):

;)

OpenStudy (solomonzelman):

\(i^2=-1\) And; \((a-b)(a+b)=a^2-b^2\) (you need to know these)

OpenStudy (lisa123):

yeah I know that you cant have complex numbers in the denominator and you have to multiply by the conjugate

OpenStudy (solomonzelman):

that is awesome at start, I see so many people that don't know that....

OpenStudy (solomonzelman):

\(\large\color{#000000 }{ \displaystyle \frac{(3-i)}{(2-i)} \times \frac{(3+i)}{(4+i)} }\) via: \((a-b)(a+b)=a^2-b^2\) and after expanding the bottom as well; we get, \(\large\color{#000000 }{ \displaystyle \frac{3^2-i^2}{8+2i-4i-i^2} }\)

OpenStudy (solomonzelman):

\(\large\color{#000000 }{ \displaystyle \frac{3^2--1}{8+6i--1} }\) \(\large\color{#000000 }{ \displaystyle \frac{9+1}{8+6i+1} }\) \(\large\color{#000000 }{ \displaystyle \frac{10}{9+6i} }\)

OpenStudy (solomonzelman):

now top and bottom times conjugae

OpenStudy (solomonzelman):

(times 9-6i)

OpenStudy (solomonzelman):

\(\large\color{#000000 }{ \displaystyle \frac{10(9-6i)}{(9+6i)(9-6i)} }\) go for it :)

OpenStudy (lisa123):

okay I think im doing it wrong but I get 90-60i/ 117

OpenStudy (solomonzelman):

Why wrong? \(\large\color{#000000 }{ \displaystyle \frac{10(9-6i)}{(9+6i)(9-6i)} }\) \(\large\color{#000000 }{ \displaystyle \frac{90-60i}{81-36i^2} }\) \(\large\color{#000000 }{ \displaystyle \frac{90-60i}{81-36(-1)} }\) \(\large\color{#000000 }{ \displaystyle \frac{90-60i}{81+36} }\) \(\large\color{#000000 }{ \displaystyle \frac{90-60i}{117} }\)

OpenStudy (solomonzelman):

You are doing it correctly, good job !!

OpenStudy (solomonzelman):

You just needed one step more to get it into an "a+bi" form

OpenStudy (solomonzelman):

\(\large\color{#000000 }{ \displaystyle \frac{90}{117}+\frac{-60}{117}i }\)

OpenStudy (solomonzelman):

1+1+7=9 Therefore 117 can be divided by 3.

OpenStudy (solomonzelman):

So now, reduce the fractions

OpenStudy (solomonzelman):

((This is another good rule, if the sum of digits of a number is divisible by 3, then the number itself is also divisible by 3.... And it is also true that if the sum of digits of a number is divisible by 9, then the number itself is divisible by 9... ))

OpenStudy (lisa123):

the answer I get is not an answer choice...

OpenStudy (lisa123):

the answer is 18/17+4/17i

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