Author Topic: Q2 TUT 0102  (Read 4706 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2607
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Q2 TUT 0102
« on: October 05, 2018, 06:12:30 PM »
Find all points of continuity of the given function:
\begin{equation*}(z)=(1-|z|^2)^{-3}.
\end{equation*}

Ye Jin

  • Full Member
  • ***
  • Posts: 17
  • Karma: 9
    • View Profile
Re: Q2 TUT 0102
« Reply #1 on: October 05, 2018, 06:54:08 PM »
Since\begin{equation*} |z|^2=x^2+y^2\end{equation*},
then  \begin{equation*}g(z)= \frac{1}{[1-(x^2+y^2)]^3}\end{equation*}   
Hence, g(z) is not continuous at all points of circle \begin{equation*}x^2+y^2=1\end{equation*}

« Last Edit: October 05, 2018, 07:09:38 PM by Ye Jin »