Complex analysis - graphing in complex plane

[jaejoon89]

Homework Statement

Graph the following in the complex plane
{zϵC: (6+i)z + (6-i)zbar + 5 = 0}

Homework Equations

z=x+iy
zbar=x-iy

The Attempt at a Solution

Substituting the equations gives
2(6x-y) + 5 = 0
=> y = 6x + (5/2)

But that's a line in R^2. The imaginary parts canceled. The question asks to graph it in the complex plane. So what will it look like?

 

[jaejoon89]

I was told that it was all complex numbers of form x + (6x + 5/2)i but i don't understand how that's derived since I got only a line in R^2

 

[Billy Bob]

Identify x+iy with the point (x,y) in the plane.

x+iy = (x,y)

Your answer is the same as the suggested answer. You found {(x,y) : y=6x +5/2}. The suggested answer is the same thing:
{x+iy : y=6x + 5/2}, i.e. {x + i(6x+5/2) : x is real}

 

[jaejoon89]

Thanks, I get the part about substituting x for y. But what does that look like in the complex plane in terms of where it crosses the real and imaginary axes?

 

[Billy Bob]

It's the line y=6x+5/2, just like in elementary algebra. The real axis is the x-axis, and the imaginary axis is the y-axis.

 

[jaejoon89]

but then isn't it an Argand diagram with coordinate axes of y and x. isn't that somewhat different than if the coordinate axes were I am and Re? (it is just somewhat confusing since in graphing it like in elementary algebra, the i is implicit otherwise it is as if it's in R^2 ! strange...)

 

[Billy Bob]

but then isn't it an Argand diagram with coordinate axes of y and x. isn't that somewhat different than if the coordinate axes were I am and Re?

No difference. That's what graphing in the complex plane means.