Find the locus of the points arg((z+1)/(z+2)) = pi

[curious__]

Homework Statement

Find the locus of the points arg((z+1)/(z+2)) = pi

Relevant Equations

z = x + iy, realising the fraction of complex numbers, using the arg = pi condition.

I tried the following proof and got -2 < x < -1 and y = 0 but my prof said that there should be something else I am missing. I have no idea what that is. Thank you.

 

[FactChecker]

I agree with your final answer but I don't think I agree with your equation for $w$. It's not immediately clear to me what you were doing there.

 

[curious__]

I agree with your final answer but I don't think I agree with your equation for $w$. It's not immediately clear to me what you were doing there.

I set z = x + iy for real numbers x and y, and then did the calculation.

 

[PeroK]

I set z = x + iy for real numbers x and y, and then did the calculation.

I don't see what's missing.

 

[fresh_42]

I couldn't find an error or another possibility either. I double-checked with WolframAlpha and it only confirmed the result.

 

[curious__]

All right. I haven't really thought of verifying with Wolfram Alpha. Yes that should confirm, I don't know, just my professor said that and I wasn't sure. Thank you anyways!

 

[FactChecker]

I set z = x + iy for real numbers x and y, and then did the calculation.

Of course. Sorry that I did not see that.
You might be interested in another way that I think is simpler:
1) Solve $w=(z+1)/(z+2)$ for $z$. That gives $z=(-2w+1)/(w-1)$.
2) Knowing that transformations of this form map straight lines and circles to straight lines and circles, map a few points along the line of arg($w$)=$\pi$ (that is $w \lt 0$). Include the endpoints of interest to see what line segment or circle arc you get. ( 0 => -1; -1/2 => (2)/(-3/2)=-4/3; -$\infty$ => -2)

Transformations $z => w$ of the type $w=(az+b)/(cz+d)$ in the complex plane are called bilinear transformations (Mobius transformations). They have nice properties that make them very convenient to use.

 

[Fred Wright]

Consider when $x=\pm \infty$