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Determine the complex number z satisfying the equation \(3z+4i\bar{z}=1-8i\)

 Jan 2, 2020
 #1
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z = 4 + 8i.

 Jan 2, 2020
 #2
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3z+4i\bar{z}=1-8i

 

\(3z+4i\bar{z}=1-8i\\ LHS=3(a+bi)+4i(a-bi)\\ LHS=3a+3bi+4ai+4b\\ LHS=3a+4b+(3b+4a)i\\ so\\ solve\;\;simultaneously\\ 3a+4b=1 \qquad and \qquad 4a+3b=-8\\ 12a+16b=4 \quad and \qquad 12a+9b=-24\\ subtract\\ 7b=28\\ b=7\\ so\\ 3a+28=1\\ 3a=-27\\ a=-9\\ so\\ a=-9\qquad b=7\\ z=-9+7i\)

 

You need to check it though.

 

Edit:

The final answer is incorrect because of a careless error near the end.   28/7 is 4

 

This error is of no significance. I am here to teach not not provide cheat sheets.

There is nothing wrong with my response. 

Especially given that I wrote it needed to be checked. But then all answers should be checked and fully understood.

 Jan 2, 2020
edited by Melody  Jan 3, 2020
 #3
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incorrect

Guest Jan 2, 2020
 #5
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to the guest above, at least say where she is incorrect. Its kind of rude to talk like that. In fact, its very rude.

CalculatorUser  Jan 3, 2020
 #6
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Thanks Guest #5

 

Atroshus is also rude for not acknowledging my answer.

Melody  Jan 3, 2020
 #7
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I'm sorry I was posting the answer I found another way as a reply to your answer.

Atroshus  Jan 3, 2020
 #8
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It is exactly the same as my answer without the little error.

I accept that you found it independently though.

 

It was a   'thank you for answering'     that I was looking for. 

 

A further comment would have been good too.

If you had bothered to look at imy answer hard enough to find my little error that would have gone down really well.

 

We do not answer these ONLY to amuse ourselves.

 

Anyway, I am glad that you know how to do them now.

Melody  Jan 3, 2020
 #4
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Just found the answer,

Let z=a+bi and \(\bar{z}=a-bi\).

so \(4i\bar{z}=4b+4ia\).

We now find that \(3z+4i\bar{z} = (3a+4b) + (4a+3b)i\)

so \(3a+4b=1\) and \(4a+3b=-8\)

Solving this leads us to a=-5 and b=4

So the answer is \(z=\boxed{-5+4i}\)

.
 Jan 2, 2020

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