Hey there Sign in to join this conversationNew here? Join for free

Differentiating complex numbers

Announcements Posted on
Study Help needs new mods! 14-04-2014
We're up for a Webby! Vote TSR to help us win. 10-04-2014
    • Thread Starter
    • 1 follower
    Offline

    ReputationRep:
    Hi
    I'm having some difficulty with this simple question:
    'Determine for what values of z (if any) the function f(z) = |z| is differentiable.'

    My attempt:
     

z=x+iy

\lvert{z}\rvert=\sqrt{x^2+y^2} \in \mathbb{R}

which is differentiable at points where x^2+y^2 >0
    Is this right? It seems too simple. Have I misunderstood the question?
    • 3 followers
    Offline

    ReputationRep:
    modulus of z is root X^2 - Y^2 as i^2 = -1

    Then I'm not sure, but I'm thinking you would differentiate implicitly by the chain rule if thats possible.
    Is this FP2?
    • 3 followers
    Offline

    (Original post by Taurus)
    modulus of z is root X^2 - Y^2 as i^2 = -1

    Then I'm not sure, but I'm thinking you would differentiate implicitly by the chain rule if thats possible.
    Is this FP2?
    That's not right, the modulus of x+iy is \sqrt{x^2+y^2}. It's the distance from the origin (in the Argand diagram) of x+iy, which corresponds to the point (x,y).

    By the looks of it this is first (maybe second) year undergraduate; differentiability definitely isn't in A-level!

    (Original post by Muffin.)
    Hi
    I'm having some difficulty with this simple question:
    'Determine for what values of z (if any) the function f(z) = |z| is differentiable.'

    My attempt:
     

z=x+iy

\lvert{z}\rvert=\sqrt{x^2+y^2} \in \mathbb{R}

which is differentiable at points where x^2+y^2 >0
    Is this right? It seems too simple. Have I misunderstood the question?
    You just seem to have stated the answer without proving it. Plug it into the definition of differentiability and take some limits to justify your claim.
    • 0 followers
    Offline

    ReputationRep:
    nooooo please dont say maths gets this hard!
    • 3 followers
    Offline

    ReputationRep:
    [QUOTE=nuodai;37162583]That's not right, the modulus of x+iy is \sqrt{x^2+y^2}. It's the distance from the origin (in the Argand diagram) of x+iy, which corresponds to the point (x,y).

    oh crap I knew that....
    • 1 follower
    Offline

    ReputationRep:
    Have you done the Cauchy-Riemann differential equations?

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?

    this is what you'll be called on TSR

  2. this can't be left blank
    this email is already registered. Forgotten your password?

    never shared and never spammed

  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By completing the slider below you agree to The Student Room's terms & conditions and site rules

  2. Slide the button to the right to create your account

    Slide to join now Processing…

    You don't slide that way? No problem.

Updated: April 15, 2012
Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.