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Mathematics 22 Online
OpenStudy (joshoyen):

Simplify. 4+2i/-6i

jimthompson5910 (jim_thompson5910):

The problem is this? \[\Large \frac{4+2i}{-6i}\]

OpenStudy (joshoyen):

Yes

jimthompson5910 (jim_thompson5910):

Multiply the fraction by \(\Large \frac{i}{i}\) \[\Large \frac{4+2i}{-6i}*\frac{i}{i} = ??\]

jimthompson5910 (jim_thompson5910):

\(\Large \frac{i}{i}\) is equal to 1. When we multiply any expression by 1, it doesn't change.

OpenStudy (joshoyen):

wait so do you multiply i/i for every problem?

jimthompson5910 (jim_thompson5910):

to turn the denominator into a real number, yes

jimthompson5910 (jim_thompson5910):

only if you have something in the form k*i in the denominator k is any real number

jimthompson5910 (jim_thompson5910):

to take advantage of the fact that i*i = i^2 = -1

OpenStudy (joshoyen):

Wait im still a bit confused whatss the next step after 4+2i/−6i ∗ i/i

jimthompson5910 (jim_thompson5910):

\[\Large \frac{4+2i}{-6i}*\frac{i}{i} =\frac{(4+2i)*i}{-6i*i} = ??\]

jimthompson5910 (jim_thompson5910):

you'll have to use the distributive property

OpenStudy (joshoyen):

so its -4-2i / 6 ?

jimthompson5910 (jim_thompson5910):

(4+2i)*i = 4i + 2i^2 = 4i+2(-1) = 4i-2

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

OpenStudy (joshoyen):

hmm okay, thanks

jimthompson5910 (jim_thompson5910):

no problem

OpenStudy (joshoyen):

How would I do 2/-6i ?

OpenStudy (joshoyen):

i'm still confused on this..

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