# Thread: [SOLVED] Image of a complex function

1. ## [SOLVED] Image of a complex function

Let $\displaystyle f(z)=z+ \frac{1}{z}$.
What's the image of :
1)A circle with radius $\displaystyle r$ centered at $\displaystyle z=0$?
2)A circular semi-ring $\displaystyle \{ z=re^{i\theta} | 0 \leq \theta \leq \pi$ , $\displaystyle 1 \leq r \leq b \}$, $\displaystyle b \geq 1$?

What's the preimage by $\displaystyle f$ of the semi-plane $\displaystyle \{ w : w \geq 0 \}$?

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For now I'm only looking for help for the 1).
My attempt : I wrote $\displaystyle z$ as $\displaystyle re^{i \theta}$ to get $\displaystyle f(re^{i\theta})=r (\cos \theta + i \sin (\theta))+ r^{-1} (\cos \theta + i \sin \theta)^{-1}$. I don't recognize what it is. Especially the part $\displaystyle r^{-1} (\cos \theta + i \sin \theta)^{-1}$.
Any idea about how I could rewrite $\displaystyle f(re^{i\theta})$?

2. Originally Posted by arbolis
Let $\displaystyle f(z)=z+ \frac{1}{z}$.
What's the image of :
1)A circle with radius $\displaystyle r$ centered at $\displaystyle z=0$?
2)A circular semi-ring $\displaystyle \{ z=re^{i\theta} | 0 \leq \theta \leq \pi$ , $\displaystyle 1 \leq r \leq b \}$, $\displaystyle b \geq 1$?
Hint: This Mobius trasformation maps $\displaystyle |z|=r$ into another circle.

What's the preimage by $\displaystyle f$ of the semi-plane $\displaystyle \{ w : w \geq 0 \}$?
Let $\displaystyle V$ be this set.

Find $\displaystyle f^{-1}$ (it exists), and note that $\displaystyle f^{-1}[V] = f^{-1}[V]$ where LHS is the inverse-image under $\displaystyle f$ and RHS is the image under $\displaystyle f^{-1}$.

3. Originally Posted by ThePerfectHacker
Hint: This Mobius trasformation maps $\displaystyle |z|=r$ into another circle.
How strange. By the way I never learned about Mobius transformations yet but I know I will in less than a month I guess.

Originally Posted by TPH
Let $\displaystyle V$ be this set.

Find $\displaystyle f^{-1}$ (it exists), and note that $\displaystyle f^{-1}[V] = f^{-1}[V]$ where LHS is the inverse-image under $\displaystyle f$ and RHS is the image under $\displaystyle f^{-1}$.
How would I find the inverse function? I'm thinking about finding a matrix and finding the inverse of it... but I'm unsure. I'd like to know how you would do it. (without posting the full solution )

4. Originally Posted by arbolis
How strange. By the way I never learned about Mobius transformations yet but I know I will in less than a month I guess.

How would I find the inverse function? I'm thinking about finding a matrix and finding the inverse of it... but I'm unsure. I'd like to know how you would do it. (without posting the full solution )
Let $\displaystyle g(z)$ be the inverse function, it means, $\displaystyle f(g(z)) = z$, therefore, $\displaystyle g(z) + \tfrac{1}{g(z)} = z$.
What does that make $\displaystyle g(z)=?$

5. Originally Posted by ThePerfectHacker
Let $\displaystyle g(z)$ be the inverse function, it means, $\displaystyle f(g(z)) = z$, therefore, $\displaystyle g(z) + \tfrac{1}{g(z)} = z$.
What does that make $\displaystyle g(z)=?$
Note sure I didn't make any mistake : $\displaystyle g(z)=\frac{z \pm \sqrt{z^2-4}}{2}$.
I should check it out.

Edit : Indeed I made an error. I'll try it again.
Edit: Ah no! It is right, at least $\displaystyle g(z)=\frac{z + \sqrt{z^2-4}}{2}$ works...
Thanks TPH, I'll try what comes next.

6. I made a mistake, I wrote "this Mobius transformation", but it is not a Mobius transformation!
So it is not necessary true that circles go to circles. Be careful in first part.

7. Originally Posted by ThePerfectHacker
I made a mistake, I wrote "this Mobius transformation", but it is not a Mobius transformation!
So it is not necessary true that circles go to circles. Be careful in first part.
Ah ok! Thanks for the correction.
It seemed strange to me because $\displaystyle f(z)=z$ would send circles to exactly the same circles, but if you add the $\displaystyle \frac{1}{z}$ term, it shouldn't be circles anymore.
A circle is mapped to the same circle plus the inverse of this circle.
I don't know what is this "inverse of this circle".

8. Hey TPH if you mind, 1) is an ellipse.