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Which region on complex plane is defined by the geometric images of $z$ that satisfied this condition:

$\frac{z}{i}-\bar{z}=0$

One of my trials was:

$\frac{z}{i}-\bar{z}=0\Leftrightarrow z- \bar{z}i=0$

If $\space z=\rho \space cis(\theta)$, and $ \space \bar{z}=\rho \space cis(-\theta)$, then

$\rho \space cis(\theta)-\rho \space cis(-\theta) \cdot cis(\frac{\pi}{2})=0 \Leftrightarrow$

$ \rho \space cis(\theta)-\rho \space cis(\frac{\pi}{2}-\theta)=0$

But, I get stuck at this stage. Even if I pass to algebric form, I can't figure it out how to solve.

Thanks for the help.

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Did you try seeing what equations $x$ and $y$ satisfy when you write $z=x+iy$ and work out the real and imaginary parts? –  Jonas Meyer May 17 '12 at 23:10

1 Answer 1

up vote 5 down vote accepted

This is one of few the occasions where writing $z=a+bi$ where $a,b$ are real numbers is the best thing you can do.

So you have $a+bi=i(a-bi)=ia+b$. Therefore $a=b$.


Another way to look at it: conjugation means symmetry with respect to the $Ox$ axis. Multiplication by $i$ means rotation by an angle of $\pi/2$ in the counterclockwise direction around the origin. $z=i\overline z$ means that if we refelct $z$ by the $x$-axis and then rotate by $\pi/2$ around the origin we get the same point. It is easy to see that the argument of $z$ can only be $\theta\in \{\pi/4,5\pi/4\}$.

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Thanks. Now I understood! I made a simulation in Geogebra to see what happens –  João May 17 '12 at 23:24

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