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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: $arg\ (z+23i)=arg\ (3\sqrt{3}i3)$ Solving: Let $z=x+jy$ $arg\ (x+2+i(y3))=arg\ (3+3\sqrt{3}i)$ $tan^{1}(\frac{y3}{x+2})=tan^{1}(\frac{\sqrt{3}}{3})$ $\frac{y3}{x+2}=\sqrt{3}$ 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 
$\dfrac{y3}{x+2} = \sqrt{3} \implies y3 = \sqrt{3}(x+2)$, 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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