Simplify quotients

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jobell

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Jul 25, 2012
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Can someone help me with this problem? I have to simplify the following quotient of complex numbers into the for
m a+bi. Show all work.
I have worked park of the problem, but do not know what to do from here...

-4-3i/2-i= (-4-3i)/(2-i) * (2+i)/(2+i)
= -8-4i-6i-3i squared/4+2i-2i-3i squared
= -8-10i-3i squared/4-2i squared
 
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jobell said:
Can someone help me with this problem? I have to simplify the following quotient of complex numbers into the for
m a+bi. Show all work.
I have worked park of the problem, but do not know what to do from here...

-4-3i/2-i= (-4-3i)/(2-i) * (2+i)/(2+1) Should be 2+i
= -8-4i-6i-3i squared/4+2i-2i-2i squared Should be -i²
= -8-10i-3i squared/4-2i squared
Click to expand...

You have a few errors in your calculation. See above. Make these changes and then let's chat if you still do not get the answer.
 
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After making the corrections, if I continue to reduce it down I get -2-10i-3i squared/-2i squared....is that what I am looking for? Thanks for all of your help!
 
jobell said:
After making the corrections, if I continue to reduce it down I get -2-10i-3i squared/-2i squared....is that what I am looking for? Thanks for all of your help!
Click to expand...

No, not correct.

\(\displaystyle \frac{(-4-3i)(2+i)}{(2-i)(2+i)}\)

\(\displaystyle \frac{-8-4i-6i-3i^2}{4+2i-2i-i^2}\)

Remember, i² = -1, therefore:

\(\displaystyle \frac{-8-10i+3}{4+1}\)

\(\displaystyle \frac{-5-10i}{5}\)

\(\displaystyle \frac{-5}{5}+\frac{-10i}{5}\)

\(\displaystyle -1-2i\)

A little tidbit here as well, in the denominator, the value will ALWAYS end up being a² + b². In your case, 2² + (-1)² = 4 + 1 = 5
 
Thank you so much!!! One day I may get this stuff right...but I doubt it.
 
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