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OpenStudy (joshoyen):
Simplify.
4+2i/-6i
10 years ago
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jimthompson5910 (jim_thompson5910):
The problem is this?
\[\Large \frac{4+2i}{-6i}\]
10 years ago
OpenStudy (joshoyen):
Yes
10 years ago
jimthompson5910 (jim_thompson5910):
Multiply the fraction by \(\Large \frac{i}{i}\)
\[\Large \frac{4+2i}{-6i}*\frac{i}{i} = ??\]
10 years ago
jimthompson5910 (jim_thompson5910):
\(\Large \frac{i}{i}\) is equal to 1. When we multiply any expression by 1, it doesn't change.
10 years ago
OpenStudy (joshoyen):
wait so do you multiply i/i for every problem?
10 years ago
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jimthompson5910 (jim_thompson5910):
to turn the denominator into a real number, yes
10 years ago
jimthompson5910 (jim_thompson5910):
only if you have something in the form k*i in the denominator
k is any real number
10 years ago
jimthompson5910 (jim_thompson5910):
to take advantage of the fact that i*i = i^2 = -1
10 years ago
OpenStudy (joshoyen):
Wait im still a bit confused whatss the next step after 4+2i/−6i ∗ i/i
10 years ago
jimthompson5910 (jim_thompson5910):
\[\Large \frac{4+2i}{-6i}*\frac{i}{i} =\frac{(4+2i)*i}{-6i*i} = ??\]
10 years ago
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jimthompson5910 (jim_thompson5910):
you'll have to use the distributive property
10 years ago
OpenStudy (joshoyen):
so its -4-2i / 6 ?
10 years ago
jimthompson5910 (jim_thompson5910):
(4+2i)*i = 4i + 2i^2 = 4i+2(-1) = 4i-2
10 years ago
jimthompson5910 (jim_thompson5910):
So we have
\[\Large \frac{4i-2}{6}\]
then we can do a bit of factoring
\[\Large \frac{4i-2}{6}=\frac{2(2i-1)}{2*3}\]
I'm sure you see what would cancel
10 years ago
OpenStudy (joshoyen):
hmm okay, thanks
10 years ago
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jimthompson5910 (jim_thompson5910):
no problem
10 years ago
OpenStudy (joshoyen):
How would I do 2/-6i ?
10 years ago
OpenStudy (joshoyen):
i'm still confused on this..
10 years ago
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