A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
prove that the function f(z) = Re(z) does not have a derivative for any value of z
anonymous
 3 years ago
prove that the function f(z) = Re(z) does not have a derivative for any value of z

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Use the limit definition of the derivative and you have it approaching different values along different lines

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how to prove it using philosophy of course very deeply??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0use CauchyRiemann conditions for differentiability. \[\newcommand {pd}{\partial}\frac{\pd u}{\pd x}=\frac{\pd v}{\pd y},\qquad \frac{\pd u}{\pd y}=\frac{\pd v}{\pd x}\] given \[f(z)=u(x,y) + v(x,y)i\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let \[z=x+yi\] \[f(z)= \text{Re } z=x+0i\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[u_x=1,v_y=0\] the first condition is not satisfied.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@sirm3d did you know symmetric tops of molecule??
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.