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I am having trouble sketching a complex argument arc

$$ \text{Sketch the following on an arcand diagram:}\\ \arg\left(\frac{w+1}{w}\right)=\frac{\pi}{6}$$

I've tried to devise a method on my own looking at questions and answers but it has failed me on this specific question so I require some help.

What I have been doing so far is I was writing $$\arg\left(\frac{x+iy+1}{x+iy}\right)$$ $$x+1=0\\y=0\\y=0 \\ x=0$$ Getting two points (-1,0) and (0,0). So I take the two points and connect them with an arc.

However the solution of the exercise was this:

enter image description here

However my answer was exactly the opposite on the y-axis. Can anyone show me a method and the logic behind the drawing of the arc, and the arc of any other similar question?

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1 Answer 1

Taking $w = x+iy$, we get that $$\text{Arg} \left( \dfrac{1+w}w\right) = \text{Arg} \left( 1+\dfrac1w\right) = \text{Arg} \left( 1+\dfrac{x-iy}{x^2+y^2}\right) = \text{Arg} \left( \dfrac{x^2+y^2+x}{x^2+y^2} - i \dfrac{y}{x^2+y^2}\right)$$ Hence, we need $$\tan(\pi/6) = -\dfrac{y}{x^2+y^2+x}$$ $$x^2 + y^2 + x + y\sqrt{3} = 0 \text{ i.e. } \left(x + \dfrac12 \right)^2 + \left(y + \dfrac{\sqrt{3}}2 \right)^2 = 1$$ However, note that we have the argument in the first quadrant and hence $$\dfrac{x^2+y^2+x}{x^2+y^2} >0 \text{ & } -\dfrac{y}{x^2+y^2} > 0$$ This gives us that $y < 0$. Hence, the curve we are interested in is $$\left(x + \dfrac12 \right)^2 + \left(y + \dfrac{\sqrt{3}}2 \right)^2 = 1 \text{ with } y< 0$$

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+1 I like your answer but it seems like too much work for the time I can spend on this part of an exercise –  εν οίδα ότι ουδέν οίδα Jun 20 '12 at 19:31

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