Toronto Math Forum
MAT3342018F => MAT334Tests => Quiz3 => Topic started by: Victor Ivrii on October 12, 2018, 06:21:48 PM

Let $0 < \alpha < 2$. Show that an appropriate choice of $\log z$ for $f(z)= z^\alpha =\exp (\alpha\log z)$ maps the domain $\{z=x + iy\colon y > 0\}$ both onetoone and onto the domain $\{w\colon 0 < \arg w < \alpha\pi\}$. Show that $f$ also carries the boundary to the boundary.
Draw both domains.

$\newcommand{\Arg}{\operatorname{Arg}}$
$\newcommand{\Log}{\operatorname{Log}}$
The range is "sector shaped," consisting of two ray boundaries. The first one lies at the positive Reaxis. The second ray starts at the origin and makes the angle $\alpha\pi$ from the first, counterclockwise. Fill in the region bounded by these two rays starting from the positive Reaxis and continue counterclockwise, and dot the border.
The range will never "overlap" as $0 < \alpha < 2$, so the possible principal arguments of $z^\alpha$ is always $0 < \Arg (z) < 2\pi$.
The appropriate choice for $\log z$ is $\Log z$.
Proving $f(z) = z^\alpha$ maps onetoone to the sector range: $f(z_1) = f(z_2) \Rightarrow z_1 = z_2$.
Let $z = re^{i\theta}$, such that $r$ is the magnitude, and $\theta$ is the principal argument. Note $0 < \theta < \pi$, the inequalities are strict.
Then $z^\alpha = r^\alpha e^{i\alpha\theta}$. Note $0 < \Arg (z^\alpha) < \alpha\pi$
Since $f(z_1) = f(z_2) = r^\alpha e^{i\alpha\theta}$, it follows that $r$ and $\theta$ for both $z_1, z_2$ are the same.
We conclude $z_1 = z_2 = re^{i\theta}$, so $f(z)$ is injective.
Proving $f(z) = z^\alpha$ maps onto the range: For all $f(z)$ on the "range" domain ${w: 0 < \Arg w < \alpha\pi}$.
Since $f(z) = r^\alpha e^{i\alpha\theta}$, it follows $z = re^{i\theta}$.
We note that $0 < \Arg f(z) < \alpha\pi$, and it follows that $0 < \Arg z < \pi$.
We conclude that $f(z)$ maps the upper halfplane onto all of the range domain.
Proving that $f$ also carries the boundary to the boundary:
The boundary of the domain domain consists of all the real numbers.
The boundary of the range domain consists of 0, the positive Reaxis, and the ray $\alpha\pi$ from that axis counterclockwise, which is $\{w: \Arg w = \alpha\pi\}$.
If $z$ is zero, $z^\alpha$ is also zero.
If $z$ is positive real, $z^\alpha$ is also positive real, with magnitude raised to the power of $\alpha$. $\Arg z = \Arg z^\alpha = 0$.
If $z$ is negative real, $\Arg z = \pi$, and it follows $\Arg z^\alpha = \alpha\pi$.The domain domain is the open upperhalf plane, the set of all the all the complex imaginary numbers with positive imaginary part. Fill the upper half of the plane and dot the border.

While domains are simple you still need to draw them