$az + b \overline{z} = 38$ is the line joining $-5 + 4i$ and $7 + 2i$. Find $(a, b)$

Question

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 $(a, b)$.

So, I decided to graph the two points out. They would be $(-5,4)$ and $(7,2)$. The slope would be $\frac{-1}{6}$. Do I substitute $z$ for $a+bi$ ?

Answer 1

You have $az+b\overline z=38$.

Let $z=x+yi.$

Then $$az+b\overline z=a(x+yi)+b(x-yi)=a(x,y)+b(x,-y)=38\dots(\star)$$

Now you just need to substitute $(-5,4)$ and $(7,2)$ in $(\star).$

And to solve the system for $a$ and $b$.

Answer 2

First of all the equation for the line with points $-5 + 4i = (-5,4)$ and $7+2i= (7,2)$ is $\frac {y - 4}{x+5} = \frac {2-4}{7+5}$ or more familiarly $y-4 = -\frac 16 x + \frac {19}6$.

But it's probably more practical to express it as $12(y-4) = -2(x+5)$. Or better yet as $2x + 12 = 38$.

That wasn't hard.

What is hard is what does "$az + b\overline z=38$ describes the line" mean?

Usually the equation $az + b\overline z = w$ will have a single solution.

That is:

If $a=m + ni = (m,n)$ and $b= j + ki = (j,k)$ and $w = u + vi=uv$ then

$az + b\overline z = w$ means

$(m,n)(x,y) + (j,k)(x,-y) = (u,v)$ or

$(mx-ny, nx+ym) + (jx + ky,kx-jy) = (u,v)$ or in other words $(x,y)$ is the point that solve

$(m+j)x +(k-n)y = u; (n+k)x + (m-j)y = v$

Or in the case where $w = 38$

$(m+j)x +(k-n)y = 38; (n+k)x + (m-j)y = 0$

Normally this one point is where the two different lines $(m+j)x +(k-n)y = 38$ and $(n+k)x + (m-j)y = 0$ intersect.

But we are told this is not a single point but a line of infinite points.

That occurs if the two equations are not linearly independent. In other words if the equations $(m+j)x +(k-n)y = 38$ and $(n+k)x + (m-j)y = 0$ are either both the same line or one is a tautology (such as $0=0$).

And it is this line:$2x + 12y = 38$.

So that means that $(m+j) = 2$ and $k-n = 12$.

However $(n+k)x + (m-j)y = 0$ can't be an equivalent line. The only way this can be mutually true is if $n+k =0$ and $m-j= 0$

So $n = -k$ and $m = j$. So $m+j = 2m = 2$ and $m = j=1$ and $k-n =2k =12$ so $k = 6; n=-6$.

So $a = (m,n) = 1 -6i$ and $b = (j,k) = 1+6i$.