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March 27th, 2017, 05:24 PM   #1
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Cartesian equation of loci from given complex equation

If I have the following formula:

$arg\ (z+2-3i)=arg\ (3\sqrt{3}i-3)$

Solving:

Let $z=x+jy$

$arg\ (x+2+i(y-3))=arg\ (-3+3\sqrt{3}i)$

$tan^{-1}(\frac{y-3}{x+2})=tan^{-1}(\frac{\sqrt{3}}{3})$

$\frac{y-3}{x+2}=-\sqrt{3}$

Is this a ray with a gradient of $-\sqrt{3}$? How do I find the starting point?
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March 28th, 2017, 08:03 AM   #2
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$\dfrac{y-3}{x+2} = -\sqrt{3} \implies y-3 = -\sqrt{3}(x+2)$, a line that passes through the point $(-2,3)$.

$\arg(-3 + i \cdot 3\sqrt{3}) \implies \theta$ is an angle in quad II $\implies \arctan(-\sqrt{3}) = \dfrac{2\pi}{3}$
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March 31st, 2017, 10:40 PM   #3
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Cheers, that clears it up. Would the slope be -1.5 or 1.5?
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