Sketching Locus of arg(z+w^*)=arg(-iw) for Complex Loci Problem

[truewt]

Homework Statement

Given that a complex number, w, is such that \pi / 2< arg(z) < \pi,

Sketch the locus described by arg(z+w^*)=arg(-iw)

Homework Equations

The Attempt at a Solution

Okay, so I attempted this question and got completely confused. I am thinking that the answer is to shade the first and fourth quadrant completely, that's an intuition (although there can be an explanation for this intuition, but I have no idea how to sketch it out online, sorry). But generally, how should I go about attempting this question? Over here, w is defined by a range instead of any distinct line, thus I am getting the problem of carrying this definition over to my solution of the problem.

 

[chaoseverlasting]

By the constraints of the problem, the locus exists in the second quadrant such that the angle it makes at any point with the x-axis is equal to the angle the complex number -iw makes with the x axis. If w is inclined at an angle \theta, then -iw is inclined to the x-axis at an angle -(\theta +\frac{\pi}{2}).

 

[truewt]

Yes, so what I was thinking is that is the answer "shade the entire first and fourth quadrant?"

 

[chaoseverlasting]

Not the fourth quadrant, because the argument is restricted from pi/2 to pi. Whoa... I wrote a whole post that's missing...basically, it came out to be a straight line segment with the slope tan(3pi/2+theta). y=tan(3pi/2+theta)x is what finally came out. This has to be restricted to the second quadrant, hence its a line segment and not a whole line. Use the eular form of i to get the angle, and since its -(theta+pi/2), the final angle is tan(pi-(-(theta+pi/2))).

 

[truewt]

Why is it only a line segment instead of an area? Theta takes values between pi and pi/2 isn't it?

 

[chaoseverlasting]

Its a line segment because the equation describes a certain family of lines with a given slope. The position of the line may vary, but that doesn't make it an area.

 

[truewt]

Doesn't the varying of the line make it an area? As in, how do you describe the locus by drawing?

 

[truewt]

By the constraints of the problem, the locus exists in the second quadrant such that the angle it makes at any point with the x-axis is equal to the angle the complex number -iw makes with the x axis. If w is inclined at an angle \theta, then -iw is inclined to the x-axis at an angle -(\theta +\frac{\pi}{2}).

are you sure its -(\theta +\frac{\pi}{2}) and not \theta -\frac{\pi}{2}?

 

[chaoseverlasting]

Yeah. I am sure. There's a minus sign to consider.