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I know that the locus of $\mathrm{arg}(z)=\theta$ is a half line with angle $\theta$, but I'm not sure why?

I can start the proof: $ z=x+iy $ $ \theta=\mathrm{arg}(z)=\arctan\left(\frac{y}{x}\right) $ $ \tan(\theta)=\frac{y}{x} $ $ y=x\cdot \tan(\theta) $ Which tells me that the locus is a line with gradient $\tan(\theta)$ passing through $(0,0)$, but I know that it should be a half line with gradient $\tan(\theta)$ starting at $(0,0)$.

Why is this?

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    I see. I only had the specific case where x > 0.2011-06-20

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Think purely in polar co-ordinates. What is the locus of complex numbers whose argument is a particular number? Pick a number like 45 degrees and try to draw it. Now add 180 degrees to this number and try again.