# Why is the **complex locus** of $\mathrm{arg}(z)=\theta$ a half line, and not a line?

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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$\arg(z)$ is not simply $\arctan(y/x)$, because that would make $\arg(1) = \arg(-1) = 0$. Rather, $\arg(z)$ is the real number $\theta$ such that $z = \lvert z \rvert \cos \theta + i \lvert z \rvert \sin \theta$. – Rahul Jun 20 '11 at 22:19
I see. I only had the specific case where x > 0. – david4dev Jun 20 '11 at 23:02

## 1 Answer

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.

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