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I have this problem that is split in 2, A and B, and Im struggling with B in particular, but I also dont know if I have done A correctly, which I suppose is necessary..

A) "Let $z$ be the complex number $z = -2 + i$ and let the angle $\phi$ be such that $tan\phi = \frac{1}{2}$ and $\frac{- \pi}{2} < \phi < \frac{\pi}{2}$.

Find the modulus $|z$ and describe the principial argument $\theta = Arg(z)$ with the help of $\phi$. Write $z$ in polar form."

So that's A.. I have found $|z| = \sqrt{(-2)^{2} + 1^{2}} = \sqrt{5}$

And the angle $\theta$ must be in quadrant II, so $ \phi = tan^{-1}(\frac{1}{2}) \approx 0.463647.. $, which means $\theta = \pi - \phi = \pi - tan^{-1}(1/2) \approx 2.678..$

Thus the polar form is: $z \approx \sqrt{5}\left(cos(2.678) + i*sin(2.678) \right)$ (Not sure if this is a good idea to write it like this with approximation..)

Hopefully I have done some what correct so far, because I really dont understand B:

"The solutions of the equation $\bar{z} = \frac{2}{z}$ can be described geometrically in the complex plane. What shape does it have?"

What I tried was to do like this:

$\bar{z} = \frac{2}{z} \Leftrightarrow z = \frac{2}{\bar{z}} $

$z = \frac{2*z}{\bar{z}*z} = \frac{2(-2+i)}{(-2-i)(-2+i)} = \frac{-4+2i}{5} = \frac{-4}{5} + \frac{2}{5}i$

I really dont know what im doing at this point. Suddenly z is something different. And even if I have done correct so far, I dont know where to go from here. How is there multiple solutions of this equation?

Hopefully I dont break any conventions on this forum, im still new here (Did search for a similiar problem first). thanks in advance

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    $\begingroup$ For your second question, you were close to answering it yourself with what you were writing. Since $ \ z \ = \ 0 \ $ cannot be a solution, we can multiply the equation through by $ \ z \ $ to obtain $ \ z \ \overline{z} \ = \ 2 \ $ , or $ \ \vert z \vert^2 \ = \ 2 \ $ . This is the set of all complex numbers with modulus $ \ \sqrt{2} \ $ , so the graph is a circle of that radius centered on $ \ z \ = \ 0 \ $. $\endgroup$ – colormegone Aug 27 '15 at 4:28
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The first part is fine. The number is in the second quadrant, has modulus $\sqrt{5}$ and argument $\pi-\arctan\frac 12$. A picture really helps.

The second part is completely unrelated. It's not the same $z$ as before.

Put $x=x+iy$ so that $\bar{z}=x-iy$ and rearrange the given equation. You will end up with the equation of a circle...

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  • $\begingroup$ Thanks! I think it was a little confusing that they used the same variable in A and B, not used to that from high school math :D $\endgroup$ – Amoz Aug 24 '15 at 15:09

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