lisa123
  • lisa123
write in the form a+bi (3-i)/(2-i) X (3+i)/(4+i)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
SolomonZelman
  • SolomonZelman
do you know what "conjugate" is?
lisa123
  • lisa123
lol I wa about to tag you
SolomonZelman
  • SolomonZelman
;)

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SolomonZelman
  • SolomonZelman
\(i^2=-1\) And; \((a-b)(a+b)=a^2-b^2\) (you need to know these)
lisa123
  • lisa123
yeah I know that you cant have complex numbers in the denominator and you have to multiply by the conjugate
SolomonZelman
  • SolomonZelman
that is awesome at start, I see so many people that don't know that....
SolomonZelman
  • 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, \(\large\color{#000000 }{ \displaystyle \frac{3^2-i^2}{8+2i-4i-i^2} }\)
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \frac{3^2--1}{8+6i--1} }\) \(\large\color{#000000 }{ \displaystyle \frac{9+1}{8+6i+1} }\) \(\large\color{#000000 }{ \displaystyle \frac{10}{9+6i} }\)
SolomonZelman
  • SolomonZelman
now top and bottom times conjugae
SolomonZelman
  • SolomonZelman
(times 9-6i)
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \frac{10(9-6i)}{(9+6i)(9-6i)} }\) go for it :)
lisa123
  • lisa123
okay I think im doing it wrong but I get 90-60i/ 117
SolomonZelman
  • SolomonZelman
Why wrong? \(\large\color{#000000 }{ \displaystyle \frac{10(9-6i)}{(9+6i)(9-6i)} }\) \(\large\color{#000000 }{ \displaystyle \frac{90-60i}{81-36i^2} }\) \(\large\color{#000000 }{ \displaystyle \frac{90-60i}{81-36(-1)} }\) \(\large\color{#000000 }{ \displaystyle \frac{90-60i}{81+36} }\) \(\large\color{#000000 }{ \displaystyle \frac{90-60i}{117} }\)
SolomonZelman
  • SolomonZelman
You are doing it correctly, good job !!
SolomonZelman
  • SolomonZelman
You just needed one step more to get it into an "a+bi" form
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \frac{90}{117}+\frac{-60}{117}i }\)
SolomonZelman
  • SolomonZelman
1+1+7=9 Therefore 117 can be divided by 3.
SolomonZelman
  • SolomonZelman
So now, reduce the fractions
SolomonZelman
  • 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... ))
lisa123
  • lisa123
the answer I get is not an answer choice...
lisa123
  • lisa123
the answer is 18/17+4/17i

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