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How do you find all complex numbers z, that satisfy the equation:

$$\arg\left(\frac{z}{z-2}\right) = \frac{\pi}{2}$$

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  • $\begingroup$ Where in the complex plane are the numbers whose argument is exactly $\pi/2$? $\endgroup$ – Umberto P. Sep 9 '15 at 15:14
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You have $$\arg\left(\frac{x+iy}{x-2+iy}\right)=\arg\left(\frac{(x+iy)(x-2-iy)}{(x-2)^2+y^2}\right)=\frac{\pi}{2}$$

The real part must be zero so$$x^2+y^2-2x=0$$ And the imaginary part must be positive, so$$y<0$$

Therefore the set of points is a semicircle of radius $1$ centre $(1, 0)$ lying below the real axis.

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Let $z = x + iy$. $$\arg\bigg(\frac{x + iy}{(x-2)+iy}\bigg) = \frac{\pi}{2}$$ Upon rationalisation, we obtain $$\arg \bigg( \frac{(x^2-2x + y^2) +i(-2y)}{(x-2)^2 + y^2} \bigg) = \frac{\pi}{2}$$ However, we know that $\frac{\pi}{2}$ lies on the imaginary axis and hence we must have that $$x^2 - 2x + y^2 = 0$$which, upon completeting the square gives $$(x-1)^2 + y^2 = 1$$ and also since it lies on the positive imaginary axis we require that $$-2y>0 \iff y<0$$

Thus we have that $z$ is any point lying in the semicircle with radius $1$ centered at $(1,0)$ that lies below the real axis.

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By using $\frac{z}{z-2}=re^{i(\frac{\pi}{2}+2n\pi)}=r$ and $z=x+iy$, we have $x+iy=r(x-2)+iry$, and thus $x=r(x-2)$ and $y=ry$.

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