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Find the locus of $\arg(\frac{z-3}{z}) = \frac{\pi}{4}$ where z represent complex number .

Working : $\arg(\frac{z-3}{z}) $ can be written as $\arg(z-3)-\arg(z) = \frac{\pi}{4}$, or $\arg\left((x-3)+iy\right) - \arg(x+iy)=\frac{\pi}{4}$.

If we take tangent to both side we get : $$ \begin{align*} \tan \left[\arg\left((x-3)+iy\right) -\arg (x+iy)\right] = \tan \frac{\pi}{4},\\ \tan\left[\frac{\arg\left((x-3)+iy\right) - \arg(x+iy)}{1+ \arg\left((x-3)+iy\right) \arg(x+iy)}\right] = 1. \end{align*} $$ Please suggest further...

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So, $$\arctan \left(\frac y{x-3}\right)-\arctan \left( \frac yx\right)=\frac\pi4$$

$$\implies \arctan\left(\frac{ \frac y{(x-3)}-\frac yx}{1+ \frac y{(x-3)}\frac yx}\right)=\frac\pi4$$

$$\implies \frac{3y}{(x-3)x+y^2}=\tan\left(\frac\pi4\right)=1$$ assuming $x(x-3)\ne0$

$$\implies x^2-3x+y^2=3y$$

If $x=3,\arctan \left(\frac y{x-3}\right)-\arctan \left( \frac yx\right)=\frac\pi4$ reduces to $$\arctan \left( \frac y0\right)-\arctan \left( \frac y3\right)=\frac\pi4$$

$$\text{Now, }\arctan \left( \frac y0\right)= \begin{cases} \ \frac{\pi}2 & \text{if } y>0 \\ \ -\frac{\pi}2 & \text{if } y<0\\ \text{ undefined} & \text{if } y= 0 \end{cases}$$

$$\text{ If }y>0, \arctan \left( \frac y3\right)=\frac\pi2-\frac\pi4=\frac\pi4$$

$$\implies \frac y3=\tan\left(\frac\pi4\right)=1\implies y=3\text{ at }x=3$$

$$\text{ If }y<0, \arctan \left( \frac y3\right)=-\frac\pi2-\frac\pi4=-\frac{3\pi}4$$

$$\implies \frac y3=\tan\left(-\frac{3\pi}4\right)=\tan\left(\pi-\frac{\pi}4\right)=-\tan\left(\frac\pi4\right)=-1\implies y=-3\text{ at }x=3$$

Similarly, for $x=0$

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$\arg(\dfrac{z-3}{z})=\dfrac{\pi}{4}\implies\Re(\dfrac{z-3}{z})=\Im(\dfrac{z-3}{z})$ and$\Re(\dfrac{z-3}{z})>0,z\neq0 $$\implies\Re(\dfrac{|z|^2-3\bar z}{|z|^2} )=\Im(\dfrac{|z|^2-3\bar z}{|z|^2})$$\implies x^2+y^2-3x=3y,y>0$(As$\Im(\dfrac{z-3}{z})=\dfrac{3y}{x^2+y^2}>0 $ if and only if $y>0$)

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z=0 should be excluded though, so this circle should have a hole at the Origin? – imranfat Mar 25 '13 at 16:53
Very nice. +1. You may want to mention that the real and imaginary parts of $\frac{z-3}{z}$ are positive because the argument is $\pi/4$, so we must have $y>0$ and the locus is not a circle but a circular arc. – user1551 Mar 25 '13 at 16:59
Didn't think of that, user1551, but that is certainly true as well. – imranfat Mar 25 '13 at 18:06

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