9

I'm confused about the straight line equation in complex plane: how does $0 = Re((m+i)z + b)$ come from $y = mx + b$? I mean when I see $y = mx + b$, I can draw a graph in my mind, but when I see $0 = Re((m+i)z + b)$, there is nothing on my mind.

How can I connect the two equatinos?

Does anyone could help me, thanks!

Martin Sleziak
  • 53,687
user133140
  • 885

3 Answers3

7

Let $z=x+iy$. Then, $(m+i)(x+iy)+b=mx+ix+imy-y+b$. Thus, looking at the real part, we get $mx-y+b=0$, also known as $y=mx+b$.

Robin Goodfellow
  • 2,821
  • I know how to prove from "0 = Re((m+i)z + b)" to "y = mx + b", but how can I do in backwards? – user133140 Sep 16 '14 at 23:12
  • 1
    @user133140: Just go in the opposite direction. Let $z=x+iy$. We have $0=mx-y+b$. Then, $0=\operatorname{Re}(mx-y+b)=\operatorname{Re}(mx-y+b+if(x,y))$, where $f$ is a real-valued function (Note that $\operatorname{Re}$ will send $if(x,y)$ to $0$). Setting $f(x,y)=x+my$, we get the desired result. – Robin Goodfellow Sep 16 '14 at 23:21
3

$mx-y+b=0\rightarrow Re(mz)+Re(iz)+Re(b)=0\rightarrow Re[(m+i)z+b]=0$

user577215664
  • 40,625
Nitin Uniyal
  • 7,946
3

When you multiply complex numbers you add the angles and multiply the lengths.

Think of $m+i$ as a vector that forms and angle of $\cot^{-1} m$ with the x axis.

The function of the line intersects the x axis at and angle of $\tan^{-1} m$.

Adding these two angles together equals $\pi/2$

When we multiply by $(m+i)$ it rotates the solutions set to a vertical line that intersects the $x-$axis at $-b.$ Adding $b,$ and it intersects at 0.

user317176
  • 11,017