My Math Forum Cartesian equation of loci from given complex equation

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 March 27th, 2017, 04:24 PM #1 Newbie   Joined: Mar 2016 From: Australia Posts: 21 Thanks: 1 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?
 March 28th, 2017, 07:03 AM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,016 Thanks: 1600 $\dfrac{y-3}{x+2} = -\sqrt{3} \implies y-3 = -\sqrt{3}(x+2)$, a line that passes through the point $(-2,3)$. $\arg(-3 + i \cdot 3\sqrt{3}) \implies \theta$ is an angle in quad II $\implies \arctan(-\sqrt{3}) = \dfrac{2\pi}{3}$
 March 31st, 2017, 09:40 PM #3 Newbie   Joined: Mar 2016 From: Australia Posts: 21 Thanks: 1 Cheers, that clears it up. Would the slope be -1.5 or 1.5?

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### obtain the cartesian equation of the loci r=atanĀ©

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