Result of a function (complex numbers)

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• Mar 7th 2011, 11:26 AM
Pranas
Result of a function (complex numbers)
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...

http://img812.imageshack.us/img812/3606/answ.jpg
• Mar 7th 2011, 11:55 AM
Opalg
Quote:

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...

http://img812.imageshack.us/img812/3606/answ.jpg

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.
• Mar 7th 2011, 12:05 PM
Pranas
Thank you! That's exactly what I was looking for
• Mar 7th 2011, 12:15 PM
Opalg
BTW the answer is particularly simple.
Spoiler:
It reduces to $|w| = 1$.