# Complex loci problem

1. Aug 20, 2007

### truewt

1. The problem statement, all variables and given/known data

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

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

2. Relevant equations

3. 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.

2. Aug 25, 2007

### 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 $$-(\theta +\frac{\pi}{2})$$.

3. Aug 26, 2007

### truewt

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

4. Aug 26, 2007

### chaoseverlasting

Not the fourth quadrant, because the argument is restricted from pi/2 to pi. Whoa... I wrote a whole post thats 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))).

5. Aug 27, 2007

### truewt

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

6. Aug 30, 2007

### 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 doesnt make it an area.

7. Aug 31, 2007

### truewt

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

8. Aug 31, 2007

### truewt

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

9. Aug 31, 2007

### chaoseverlasting

Yeah. Im sure. There's a minus sign to consider.