Rays and Lines in the Complex Plane

Here we are going to cover some basics of sketching in the complex plane. 

$\left\{z:arg\left(z\right)=\theta\right\}${z:arg(z)=θ} is a ray starting at origin with an angle of $\theta$θ with the positive real axis.

 

$\left\{z:arg\left(z+a+bi\right)=\theta\right\}${z:arg(z+a+bi)=θ} is a ray with an angle of $\theta$θ with the positive real axis translated from the origin a units in real direction and $b$b units in the imaginary direction. 

 

$\left\{z:Re\left(z\right)=a\right\}${z:Re(z)=a} for $a$a ∈ ℝ is a vertical line through $x=a$x=a.

$\left\{z:Re\left(z+a+bi\right)=c\right\}${z:Re(z+a+bi)=c} for $a,b,c$a,b,c ∈ ℝ is a vertical line translated ($-a$−a) in the Real direction and ($-b$−b) in the Imaginary direction. Since this graph is a vertical line the imaginary translation is irrelevant. This is a vertical line passing through $\left(c-a\right)+0i$(c−a)+0i

 

$\left\{z:Im\left(z\right)=c\right\}${z:Im(z)=c} for $c$c ∈ ℝ is a horizontal line through $y=c$y=c.

 

$\left\{z:Im\left(z+a+bi\right)=c\right\}${z:Im(z+a+bi)=c} for $a,b,c$a,b,c ∈ ℝ is a horizontal line translated ($-a$−a) in the Real direction and ($-b$−b) in the Imaginary direction. Since this graph is a horizontal line the imaginary real component is irrelevant. This is a horizontal line passing through $0+\left(c-b\right)i$0+(c−b)i

 

If $z=x+iy$z=x+iy, then $z+a+bi=c$z+a+bi=c becomes $x+iy+a+bi=c$x+iy+a+bi=c, which is $\left[x+a\right]+\left[y+bi\right]=c$[x+a]+[y+bi]=c. So then if we want $Re\left(z+a+bi\right)=c$Re(z+a+bi)=c then we just look at $x+a=c$x+a=c, and if we wanted the $Im\left(z+a+bi\right)=c$Im(z+a+bi)=c then we just look at $y+b=c$y+b=c.

Worked examples

Example 1

Sketch $6=Re\left[\left(2-3i\right)z\right]$6=Re[(2−3i)z]

First we let $z=x+iy$z=x+iy

Then

$\left(2-3i\right)z$(2−3i)z	$=$=	$\left(2-3i\right)\left(x+iy\right)$(2−3i)(x+iy)
	$=$=	$2x-3ix+2iy+3y$2x−3ix+2iy+3y
	$=$=	$\left(2x+3y\right)+\left(2x-3x\right)i$(2x+3y)+(2x−3x)i

 

So, when we want to sketch $6=Re\left[\left(2-3i\right)z\right]$6=Re[(2−3i)z], (the REAL COMPONENT), we need only look at the real component of $\left(2x+3y\right)+\left(2x-3x\right)i$(2x+3y)+(2x−3x)i, which is $\left(2x+3y\right)$(2x+3y).

$Re\left[\left(2-3i\right)z\right]=6$Re[(2−3i)z]=6 is 

$\left(2x+3y\right)=6$(2x+3y)=6

 

Example 2

Sketch $-2=Im\left[\left(4+2i\right)z\right]$−2=Im[(4+2i)z]

Let

$z=x+iy$z=x+iy, then 

$\left(4+2i\right)z$(4+2i)z	$=$=	$\left(4+2i\right)\left(x+iy\right)$(4+2i)(x+iy)
	$=$=	$4x+2ix+4iy-2y$4x+2ix+4iy−2y
	$=$=	$\left(4x-2y\right)+\left(2x+4y\right)i$(4x−2y)+(2x+4y)i

So, when we want to sketch $-2=Im\left[\left(4+2i\right)z\right]$−2=Im[(4+2i)z], (the IMAGINARY COMPONENT), we need only look at the imaginary component of $\left(4x-2y\right)+\left(2x+4y\right)i$(4x−2y)+(2x+4y)i, which is $\left(2x+4y\right)$(2x+4y).

$Im\left[\left(4+2i\right)z\right]=-2$Im[(4+2i)z]=−2 is 

$\left(2x+4y\right)=-2$(2x+4y)=−2

 

Have a play with this applet, it will allow you to sketch a variety of lines on the complex plane. 

 

Example 3

Sketch

$28=Im\left[\left(9-i\right)z-5\left(2+2i\right)z\right]$28=Im[(9−i)z−5(2+2i)z]

Let $z=x+iy$z=x+iy

$\left(9-i\right)z-5\left(2+2i\right)z$(9−i)z−5(2+2i)z	$=$=	$\left(9-i\right)\left(x+iy\right)-5\left(2+2i\right)\left(x+iy\right)$(9−i)(x+iy)−5(2+2i)(x+iy)
	$=$=	$9x+9iy-ix+y-5\left(2x+2ix+2iy-2y\right)$9x+9iy−ix+y−5(2x+2ix+2iy−2y)
	$=$=	$9x+9iy-ix+y-10x+10ix+10iy-10y$9x+9iy−ix+y−10x+10ix+10iy−10y
	$=$=	$9x-10x+y-10y-ix+10ix+10iy+9iy$9x−10x+y−10y−ix+10ix+10iy+9iy
	$=$=	$\left(-x-9y\right)+\left(-11x-19y\right)i$(−x−9y)+(−11x−19y)i

So to sketch, we are only interested in the imaginary component, 

$-11x-19y=28$−11x−19y=28