# mapping a sector under a complex function

Help needed below with mapping an area of sector

Consider the function, $z\in \mathbb{C}$ $$u+iv=f(z)=f(x+iy)=\frac{1}{2} \Big(z+\frac{1}{z}\Big)$$

1. Determine the domain of analyticity of $f(z)$

For this use Cauchy-Reimann $u_x=v_y\,$, $u_y=-v_x$ to get $y^2=x^2\implies y=\pm x$

1. Find the image of the locus $|z|=R$ discussing the cases $R>1$, $R<1$ and $R=1$

For $R=1$, $z=\cos\theta$, so it maps onto the line $z\in [-1,1]$

Substitute $z=r(\cos\theta+i\sin\theta)$ to get $w=u+iv$ $$u=\frac{1}{2} (r+\frac{1}{r})\cos\theta \\ v=\frac{1}{2} (r-\frac{1}{r})\sin\theta$$ which I think looks like an ellipse $$\frac{u^2}{(r+1/r)^2}+\frac{v^2}{(r-1/r)^2}=\frac{1}{2}^2$$

Not sure about where the ellipse is outside and where it is inside the unit circle for the cases $R\neq1$.

1. Find the image of the finite region of the complex plane bounded by the curves $z_1=\rho$, $z_2=\rho e^{i\frac{\pi}{4}}$, with $0\leq \rho \leq 1$, and $|z_3|=1$

I understand this is a $45$ degrees slice from a unit circle, and that it maps onto and ellipse, but not sure exactly how, and what is the most efficient way to do this - but I am thinking that the image will be un-bounded sector.

UPDATE For 3. I think the image is the inside of a hyperbola as shaded below - but still not entirely sure I am correct:

The hyberbolic equation $(u/\cos\theta)^2-(v/\sin\theta)^2=4$ is attained from eliminating $r$ from the equations for $u$ and $v$ above.

1. Determine the points for which the map $f(z)$ is not conformal; compute the factor by which angles between the tangents to two lines outgoing from such points get multiplied under the map

$$f(z)=\frac{1}{2}(z+\frac{1}{z})\\ f'(z)=\frac{1}{2}(1-\frac{1}{z^2})=0\\ \implies z=\pm 1$$

Not sure how best to compute the factor for the angles

1. For a generic analytic function $g(z)$, find the condition under which the angles between the tangents to two lines outgoing from a point $z$ get multiplied by a factor $m\in \mathbb{N}$ under the map

• For 1. I think it is analytic on $\mathbb{C}\backslash\{0\}$. – Nosrati Sep 6 '17 at 18:12
• How do you compute $u_x$.? – Nosrati Sep 6 '17 at 18:20
• @MyGlasses from the function $f(x+iy)=u+iv; \, u,v,x,y\in\mathbb{R}$ (sorry I should mention this above, will edit it) and then partially differentiate $u$ wrt $x$ – Zeeshan Ahmad Sep 6 '17 at 18:34