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ganeshie8

  • one year ago

\[\large \dfrac{d|z|}{dz} = ?\] \(z\) is a complex number

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  1. UsukiDoll
    • one year ago
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    z is a complex number? like z =ai+b ? a is the real part and b is the imaginary part. hmm I'm thinking in terms of Euclidean Geometry , but I don't think it applies here :/

  2. ganeshie8
    • one year ago
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    Yes \(z = a+ib\) then \(|z| = \sqrt{a^2+b^2}\)

  3. UsukiDoll
    • one year ago
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    that's where I die... it's like I find the definitions first and then it seems that we have to find the derivative of absolute value of z which is a complex number. z = a+bi. Then my brain explodes.

  4. UsukiDoll
    • one year ago
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    hmmm \[\mid z \mid = \sqrt{a^2+b^2}\] \[\mid z \mid = (a^2+b^2)^\frac{1}{2}\] then find derivative ? let a and b be constants?

  5. ganeshie8
    • one year ago
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    \[\dfrac{d|z|}{dz} = \lim\limits_{\Delta z\to 0} \dfrac{|z+\Delta z| - |z|}{\Delta z}\] Let \(z=x+iy\), then the above is is same as \[\dfrac{d|z|}{dz} = \lim\limits_{(\Delta x\to 0, ~\Delta y\to 0)} \dfrac{|(x+\Delta x)+i(y+\Delta y)| - |x+iy|}{\Delta x + i\Delta y}\]

  6. sparrow2
    • one year ago
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    if a nad b are constant then sqrt(a^2+b^2) will be constant too. so it doesn't change and i think derivative is 0 or i miss smth :)

  7. ganeshie8
    • one year ago
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    If \(z=a+ib\), then \(a\) and \(b\) change as \(z\) changes, so they are not really constants with respect to \(z\) here

  8. UsukiDoll
    • one year ago
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    oh my O_O! so a and b is connected to z then

  9. sdfgsdfgs
    • one year ago
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    My GUESS is that it is undefined. First, consider d|x|/dx = 1 or -1 depending on x is > or < 0 Now, for z, consider it in polar form (r, theta), |z|=r, so \[\frac{ |z+ \Delta z| - |z| }{ \Delta z } = \frac{ r + \Delta r - r }{ \Delta z } = \frac{ \Delta r }{ \Delta z }\] if z is a complex variable with only iy component (x=0), then delta r / delta z = 1 or -1 depending on y is > or <0 but for any other complex no, delta r / delta z will be a complex no. depending on the theta component. The influence of the theta component may not be eliminated by the limit delta z approaches 0. so i think d|z|/dz is undefined...just my guess only though.

  10. freckles
    • one year ago
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    what if we think of a and b as real-valued functions of x then z is too \[|z|(x)=\sqrt{[a(x)]^2+[b(x)]^2} \\ |z|'(x)=\frac{1}{2 \sqrt{a^2(x)+b^2(x)}} (2a(x)a'(x)+2 b(x) b'(x)) \\ |z|'(x)=\frac{aa'+bb'}{\sqrt{a^2+b^2}} \\ \text{ So we have } \\ \frac{d}{dx}|z|=\frac{aa'+bb'}{\sqrt{a^2+b^2}} \\ d|z|=\frac{aa'+bb'}{\sqrt{a^2+b^2}}dx \\ \frac{d|z|}{dz}=\frac{aa'+bb'}{\sqrt{a^2+b^2}} \frac{dx}{dz} \\ z(x)=a(x)+b(x) \cdot i \\ \frac{dz}{dx}=a'+b' \cdot i \\ \frac{d|z|}{dz}=\frac{aa'+bb'}{\sqrt{a^2+b^2}} \frac{1}{a'+b' i}\]

  11. ganeshie8
    • one year ago
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    @sdfgsdfgs why am i not allowed to replace \(\Delta z\) by \(\Delta r\) and get \[\frac{ |z+ \Delta z| - |z| }{ \Delta z } = \frac{ r + \Delta r - r }{ \Delta r } = \frac{ \Delta r }{ \Delta r} = 1\] ?

  12. freckles
    • one year ago
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    let me know if tht above is like really really wrong or whatever

  13. sdfgsdfgs
    • one year ago
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    @ganeshie8 hahahaa thats what my FIRST guess as well :) then i realize delta z is a complex no. so after the division, delta r / delta z will remain a complex no.

  14. ganeshie8
    • one year ago
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    Ohkk.. so I think it all looks good, we can conlcude the derivative doesn't exist from freckles work too as \(a'\) and \(b'\) depend on the direction in which we approach \(z\)

  15. ganeshie8
    • one year ago
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    For the limit of a function of two variables to exist, the limit must exist in all directions and be the same

  16. freckles
    • one year ago
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    so we are saying \[\frac{d|z|}{dz}=\frac{aa'+bb'}{\sqrt{a^2+b^2}} \frac{1}{a'+b' i} \text{ doesn't exist } ? \\ \text{ so } z=2x -5i \\ \text{ so } \frac{d|z|}{dz} \text{ doesn't exist } \\ \text{ or is \it } \frac{4x}{\sqrt{4x^2+25}} \frac{1}{2-0i} =\frac{2x}{\sqrt{4x^2+25} }\]

  17. ganeshie8
    • one year ago
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    I think we're saying \(\dfrac{d|z|}{dz}\) exists in a specific direction but it is not the same in all directions

  18. ganeshie8
    • one year ago
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    |dw:1436697803754:dw|

  19. ganeshie8
    • one year ago
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    \(f'(3+i2)\) in each direction need not be same, so the derivative doesn't exist

  20. ganeshie8
    • one year ago
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    Wow! how did you get that wil try to figure it out... have good sleep :)

  21. sdfgsdfgs
    • one year ago
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    @ganeshie8 think about delta r / delta z n what u just posted about the "direction" (which is theta...

  22. sdfgsdfgs
    • one year ago
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    jump out of bed to type this :D d|z|/dz = cos(theta) + i*sin (theta) where z=(r, theata) im really going back to bed now!

  23. UsukiDoll
    • one year ago
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    \[\frac{d \mid z \mid}{dz} = \cos(\theta) +isin(\theta), z =(r, \theta) \]

  24. ganeshie8
    • one year ago
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    \[\large \frac{d|z|}{dz} = e^{i\arg(z)}\] ? this looks interesting

  25. ytrewqmiswi
    • one year ago
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    we define dy/dx as rate of change of y with respect to x. in this case dIzI/dz would mean rate of change of magnitude of z with respect to z. we know that we can't compare any complex number with any other number so this implies that we can't even find the RATE of change of IzI with respect to z.

  26. ganeshie8
    • one year ago
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    In derivatives/limits we're dealing with absolute values of complex numbers, which are real. So it should be fine ?

  27. freckles
    • one year ago
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    http://mathworld.wolfram.com/AbsoluteValue.html

  28. freckles
    • one year ago
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    d|z|/dz doesn't exist according to this giving an explanation about direction as @ganeshie8 was saying

  29. sdfgsdfgs
    • one year ago
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    @freckles is right should have stayed in bed - my last post ignored delta theta. in evaluating delta r / delta z, both theta AND delta theta needs to be consider. So d|z|/dz should be undefined. sorry for the confusion!

  30. ganeshie8
    • one year ago
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    I'm going to work the derivatives in \(x=0\), \(y=0\) and \(y=mx\) directions and see if they are different

  31. freckles
    • one year ago
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    http://www.wolframalpha.com/input/?i=d%7Cz%7C%2Fdz%2C+z+is+complex @ganeshie8 how do I interpret this? It says it is 0 but there are no complex solutions.

  32. freckles
    • one year ago
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    if you know anyways (I know you are the wolfram master)

  33. ganeshie8
    • one year ago
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    Looks there is an issue with the input interpretation, wolf is treating the input as \[\text{solve}~~\dfrac{d|z|}{dz} = 0~~\text{over complex numbers}\]

  34. freckles
    • one year ago
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    oh

  35. ganeshie8
    • one year ago
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    As expected it says there are no solutions

  36. freckles
    • one year ago
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    I completely misread that

  37. ytrewqmiswi
    • one year ago
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    @ganeshie8 in this case we have taken only one absolute value(IzI) nd still there is a complex number (z) due to which u can't differentiate

  38. freckles
    • one year ago
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    I thought it was saying at first the derivative of |z| w.r.t to z was 0 but yes I get that is how it interpreted my question :( to solve

  39. ganeshie8
    • one year ago
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    \[\dfrac{d|z|}{dz} \Bigg|_{\text{over circle}} = 0\] ?

  40. ytrewqmiswi
    • one year ago
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    ?? what does that over circle mean??

  41. ytrewqmiswi
    • one year ago
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    integration?

  42. freckles
    • one year ago
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    http://www.wolframalpha.com/input/?i=limit%28%28%7Cz%2Bh%7C-%7Cz%7C%29%2Fh%2Ch%3D0%29

  43. freckles
    • one year ago
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    oh but I think h there is meaning change in real number only

  44. ganeshie8
    • one year ago
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    nvm, i was thinking stupid something like staying on a circle around 3+i2 for finding f'(3+i2) |dw:1436699672546:dw|

  45. ytrewqmiswi
    • one year ago
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    i have i more reason which proves that u can't do differentiation with complex numbers

  46. ganeshie8
    • one year ago
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    Yes, @freckles if we are approaching \(3+i2\) along real axis, then \(\Delta y = 0\) : \[\begin{align}\dfrac{d|z|}{dz}\Bigg|_{\Delta y=0} &= \lim\limits_{(\Delta x\to 0, ~\Delta y = 0)} \dfrac{|(x+\Delta x)+i(y+0)| - |x+iy|}{\Delta x + i0}\\~\\ &= \lim\limits_{(\Delta x\to 0, ~\Delta y = 0)} \dfrac{|(x+\Delta x)+i(y+0)| - |x+iy|}{\Delta x } \\~\\ &= \dfrac{x}{\sqrt{x^2+y^2}}\\~\\ &=\dfrac{x}{|z|} \end{align}\]

  47. ganeshie8
    • one year ago
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    If we are approaching \(3+i2\) along imaginary axis, then \(\Delta x = 0\) : \[\begin{align}\dfrac{d|z|}{dz}\Bigg|_{\Delta x=0} &= \lim\limits_{(\Delta x=0, ~\Delta y \to 0)} \dfrac{|(x+ 0 )+i(y+\Delta y)| - |x+iy|}{0 + i\Delta y}\\~\\ &= \frac{1}{i}\lim\limits_{(\Delta x=0, ~\Delta y \to 0)} \dfrac{|x+i(y+\Delta y)| - |x+iy|}{\Delta y } \\~\\ &= \dfrac{-iy}{\sqrt{x^2+y^2}}\\~\\ &=-\dfrac{iy}{|z|} \end{align}\]

  48. ytrewqmiswi
    • one year ago
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    reason - Ok now we have dIzI/dz first of all we need to know that there exist more than 1 right angled triangles with same hypotenuse . now we have complex number z=b+ai IzI=(b^2+a^2)^(1/2) remember to get c^2 = a^2+b^2 there are more than 1 pairs of a and b which satisfy. so here IzI can be obtained by 2 different complex numbers. so we have d(b^2+a^2)^(1/2) /d(b+ai) now we see that the numerator part is constant i.e, it can have only 1 value but as stated above we can have 2 or more than 2 different complex numbers which can have mag. = z so we can have dIzI/dz = dIzI/dx = dIzI/dq ....and so one where x,q,....are compex numbers with mag.=IzI but we can't have dy/dx = dy/dz where x and z are different so this proves that u can't differentiate wrt a complex number

  49. ganeshie8
    • one year ago
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    That looks interesting, are you saying \(|z|^2 = a^2+b^2\) having multiple solutions is somehow bad for differentiability of \(f(z) = |z|\) ?

  50. ytrewqmiswi
    • one year ago
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    yes

  51. freckles
    • one year ago
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    @ganeshie8 have you ever head of something called cauchy riemann equation?

  52. freckles
    • one year ago
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    heard*

  53. freckles
    • one year ago
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    like I did something above with a and b being in terms of x i might be misunderstand maybe I'm suppose to consider a and b over two variables instead of one

  54. freckles
    • one year ago
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    since you know z is complex

  55. ganeshie8
    • one year ago
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    \[f(x+iy) = u(x,y) + iv(x,y)\] \(f\) is differentiable if and only if \[u_x = v_y~~~\land~~~u_y=-v_x\]

  56. ytrewqmiswi
    • one year ago
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    Holomorphic Function?

  57. ganeshie8
    • one year ago
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    that works only for functions that output complex numbers i think our function f(z) = |z| is a real valued function, so idk if we can use cauchy riemann condition

  58. freckles
    • one year ago
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    i think you are right

  59. ganeshie8
    • one year ago
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    If it does work, then we get a pretty interesting result : The derivative of any real valued complex function, if it exists, is \(0\).

  60. ganeshie8
    • one year ago
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    Because for a real valued function, we have \(v(x,y) = 0\) \[f(x+iy) = u(x,y) + i0\] \(u_x = v_y = 0~~\land u_y = -v_x = 0\) imply that \(\dfrac{df}{fz} = 0\) for all real valued differentiable complex functions

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