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The equation of the line joining the complex numbers -5 + 4i and 7 + 2i can be expressed in the form az + b \overline{z} = 38 for some complex numbers a and b. Find the ordered pair (a, b).

 May 4, 2019
edited by Guest  May 4, 2019
edited by Guest  May 4, 2019
 #1
avatar+6046 
+3

\(\text{Both points are on the line so plug them in}\\ a(-5+4i) + b(-5-4i)=38\\ a(7+2i)+b(7-2i) = 38\\ (-5a-5b)+i(4a-4b)=38\\ (7a+7b)+i(2a-2b)=38\)

 

\(-5(a+b)-14(a+b)=-38\\ -19(a+b)=-38\\ a+b=2\)

 

\(-5(2) + i(4a-4(2-a))=38\\ i(8a-8)=48\\ i(a-1)=6\\ a-1 = -6i\\ a=1-6i\\ b=2-(1-6i) = 1+6i\)

 

\((a,b) = (1-6i,1+6i)\)

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 May 4, 2019
 #2
avatar+107085 
0

Hi Rom,

Thanks for your answer  laugh

 

but...

I do not understand that initial plug in.  Why have you used the conjugates??

Could you explain a little more please. 

 May 5, 2019
 #3
avatar+6046 
+1

we're told that both points lie on the line

 

\(a z + b \bar{z} = 38\)

 

\(\text{So just plug both of these points into the formula there and we solve for }a \text{ and }b\\ \text{I use the complex conjugate because that's how they've specified the line}\)

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 May 5, 2019
edited by Rom  May 5, 2019
 #4
avatar+107085 
0

Thanks,

I wasn't twigging with the \overline z bit.  I just didn't think about the Latex conversion properly, it comes out on a new line for me so I overlooked it.  frown

Melody  May 5, 2019

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