Result of a function (complex numbers)

**Pranas** · March 7th 2011, 11:26 AM

I need to define the possible outcome of a function

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

whereas

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

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

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

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

I find it fairly complicated, because while computer graphs it nicely as $\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...

**Opalg** · March 7th 2011, 11:55 AM

Originally Posted by Pranas

I need to define the possible outcome of a function

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

whereas

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

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

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

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

I find it fairly complicated, because while computer graphs it nicely as $\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: $z = \dfrac{-3w}{w-2i}.$

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

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

**Pranas** · March 7th 2011, 12:05 PM

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

**Opalg** · March 7th 2011, 12:15 PM

BTW the answer is particularly simple.

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