# Thread: Result of a function (complex numbers)

1. ## Result of a function (complex numbers)

I need to define the possible outcome of a function

$\displaystyle \displaystyle $\omega \left( {\rm Z} \right) = \frac{{2i \cdot {\rm Z}}}{{{\rm Z} + 3}}$$

whereas

$\displaystyle \displaystyle $\omega = u + i \cdot v \in \mathbb{C}$$

$\displaystyle \displaystyle ${\rm Z} = x + i \cdot y \in \mathbb{C}$$

and my $\displaystyle \displaystyle ${\rm Z}$$ obeys the equality

$\displaystyle \displaystyle $\left| {{\rm Z} - 1} \right| = 2$$

I find it fairly complicated, because while computer graphs it nicely as $\displaystyle \displaystyle $\left| {u + i \cdot v} \right| = 1$$ , my writings (fragment added below) and squared parts of equation were not going to be nearly as easily interpreted.

Maybe you know a smart and elegant way? It was not supposed to be a very complicated problem...

2. Originally Posted by Pranas
I need to define the possible outcome of a function

$\displaystyle \displaystyle $\omega \left( {\rm Z} \right) = \frac{{2i \cdot {\rm Z}}}{{{\rm Z} + 3}}$$

whereas

$\displaystyle \displaystyle $\omega = u + i \cdot v \in \mathbb{C}$$

$\displaystyle \displaystyle ${\rm Z} = x + i \cdot y \in \mathbb{C}$$

and my $\displaystyle \displaystyle ${\rm Z}$$ obeys the equality

$\displaystyle \displaystyle $\left| {{\rm Z} - 1} \right| = 2$$

I find it fairly complicated, because while computer graphs it nicely as $\displaystyle \displaystyle $\left| {u + i \cdot v} \right| = 1$$ , my writings (fragment added below) and squared parts of equation were not going to be nearly as easily interpreted.

Maybe you know a smart and elegant way? It was not supposed to be a very complicated problem...

Start from the third line of your solution: $\displaystyle z = \dfrac{-3w}{w-2i}.$

Then $\displaystyle z-1 = \dfrac{-4w+2i}{w-2i}$, so $\displaystyle |z-1| = \dfrac{|-4w+2i|}{|w-2i|} = 2.$ Hence $\displaystyle |-2w+i| = |w-2i|.$

Therefore $\displaystyle |-2w+i|^2 = |w-2i|^2$. Now use the fact that if $\displaystyle \omega = \xi+i\eta$ then $\displaystyle |\omega|^2 = \xi^2 + \eta^2$, to see that $\displaystyle (-2u)^2 + (-2v+1)^2 = u^2 + (v-2)^2$. Simplify that to get the equation of a circle.

3. Thank you! That's exactly what I was looking for

4. BTW the answer is particularly simple.
Spoiler:
It reduces to $\displaystyle |w| = 1$.