Author Topic: Q3 TUT 5301  (Read 701 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2447
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Q3 TUT 5301
« 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 one-to-one and onto the domain $\{w\colon 0 < \arg w < \alpha\pi\}$. Show that $f$ also carries the boundary to the boundary.

Draw both domains.

oighea

  • Full Member
  • ***
  • Posts: 19
  • Karma: 21
  • I am a mulligan!
    • View Profile
Re: Q3 TUT 5301
« Reply #1 on: October 12, 2018, 06:52:26 PM »
$\newcommand{\Arg}{\operatorname{Arg}}$
$\newcommand{\Log}{\operatorname{Log}}$
The range is "sector shaped," consisting of two ray boundaries. The first one lies at the positive Re-axis. 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 Re-axis 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 one-to-one 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 half-plane 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 Re-axis, 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 upper-half 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.
« Last Edit: November 01, 2018, 08:41:51 PM by oighea »

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2447
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Q3 TUT 5301
« Reply #2 on: October 14, 2018, 07:11:11 AM »
While domains are simple you still need to draw them