+0

# help would be great

0
71
4

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
+5226
+3

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

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

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

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

.
May 4, 2019
#2
+102461
0

Hi Rom,

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
+5226
+1

we're told that both points lie on the line

$$a z + b \bar{z} = 38$$

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

.
May 5, 2019
edited by Rom  May 5, 2019
#4
+102461
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.

Melody  May 5, 2019