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ganeshie8
 one year ago
\[\large \dfrac{dz}{dz} = ?\]
\(z\) is a complex number
ganeshie8
 one year ago
\[\large \dfrac{dz}{dz} = ?\] \(z\) is a complex number

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UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1z 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 :/

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Yes \(z = a+ib\) then \(z = \sqrt{a^2+b^2}\)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1that'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.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1hmmm \[\mid z \mid = \sqrt{a^2+b^2}\] \[\mid z \mid = (a^2+b^2)^\frac{1}{2}\] then find derivative ? let a and b be constants?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[\dfrac{dz}{dz} = \lim\limits_{\Delta z\to 0} \dfrac{z+\Delta z  z}{\Delta z}\] Let \(z=x+iy\), then the above is is same as \[\dfrac{dz}{dz} = \lim\limits_{(\Delta x\to 0, ~\Delta y\to 0)} \dfrac{(x+\Delta x)+i(y+\Delta y)  x+iy}{\Delta x + i\Delta y}\]

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.1if 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 :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4If \(z=a+ib\), then \(a\) and \(b\) change as \(z\) changes, so they are not really constants with respect to \(z\) here

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1oh my O_O! so a and b is connected to z then

sdfgsdfgs
 one year ago
Best ResponseYou've already chosen the best response.1My GUESS is that it is undefined. First, consider dx/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 dz/dz is undefined...just my guess only though.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1what if we think of a and b as realvalued functions of x then z is too \[z(x)=\sqrt{[a(x)]^2+[b(x)]^2} \\ z'(x)=\frac{1}{2 \sqrt{a^2(x)+b^2(x)}} (2a(x)a'(x)+2 b(x) b'(x)) \\ z'(x)=\frac{aa'+bb'}{\sqrt{a^2+b^2}} \\ \text{ So we have } \\ \frac{d}{dx}z=\frac{aa'+bb'}{\sqrt{a^2+b^2}} \\ dz=\frac{aa'+bb'}{\sqrt{a^2+b^2}}dx \\ \frac{dz}{dz}=\frac{aa'+bb'}{\sqrt{a^2+b^2}} \frac{dx}{dz} \\ z(x)=a(x)+b(x) \cdot i \\ \frac{dz}{dx}=a'+b' \cdot i \\ \frac{dz}{dz}=\frac{aa'+bb'}{\sqrt{a^2+b^2}} \frac{1}{a'+b' i}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4@sdfgsdfgs why am i not allowed to replace \(\Delta z\) by \(\Delta r\) and get \[\frac{ z+ \Delta z  z }{ \Delta z } = \frac{ r + \Delta r  r }{ \Delta r } = \frac{ \Delta r }{ \Delta r} = 1\] ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1let me know if tht above is like really really wrong or whatever

sdfgsdfgs
 one year ago
Best ResponseYou've already chosen the best response.1@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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Ohkk.. 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 \(z\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4For the limit of a function of two variables to exist, the limit must exist in all directions and be the same

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so we are saying \[\frac{dz}{dz}=\frac{aa'+bb'}{\sqrt{a^2+b^2}} \frac{1}{a'+b' i} \text{ doesn't exist } ? \\ \text{ so } z=2x 5i \\ \text{ so } \frac{dz}{dz} \text{ doesn't exist } \\ \text{ or is \it } \frac{4x}{\sqrt{4x^2+25}} \frac{1}{20i} =\frac{2x}{\sqrt{4x^2+25} }\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I think we're saying \(\dfrac{dz}{dz}\) exists in a specific direction but it is not the same in all directions

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1436697803754:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\(f'(3+i2)\) in each direction need not be same, so the derivative doesn't exist

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Wow! how did you get that wil try to figure it out... have good sleep :)

sdfgsdfgs
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 think about delta r / delta z n what u just posted about the "direction" (which is theta...

sdfgsdfgs
 one year ago
Best ResponseYou've already chosen the best response.1jump out of bed to type this :D dz/dz = cos(theta) + i*sin (theta) where z=(r, theata) im really going back to bed now!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{d \mid z \mid}{dz} = \cos(\theta) +isin(\theta), z =(r, \theta) \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[\large \frac{dz}{dz} = e^{i\arg(z)}\] ? this looks interesting

ytrewqmiswi
 one year ago
Best ResponseYou've already chosen the best response.1we 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4In derivatives/limits we're dealing with absolute values of complex numbers, which are real. So it should be fine ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1dz/dz doesn't exist according to this giving an explanation about direction as @ganeshie8 was saying

sdfgsdfgs
 one year ago
Best ResponseYou've already chosen the best response.1@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 dz/dz should be undefined. sorry for the confusion!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I'm going to work the derivatives in \(x=0\), \(y=0\) and \(y=mx\) directions and see if they are different

freckles
 one year ago
Best ResponseYou've already chosen the best response.1http://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.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1if you know anyways (I know you are the wolfram master)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Looks there is an issue with the input interpretation, wolf is treating the input as \[\text{solve}~~\dfrac{dz}{dz} = 0~~\text{over complex numbers}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4As expected it says there are no solutions

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I completely misread that

ytrewqmiswi
 one year ago
Best ResponseYou've already chosen the best response.1@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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[\dfrac{dz}{dz} \Bigg_{\text{over circle}} = 0\] ?

ytrewqmiswi
 one year ago
Best ResponseYou've already chosen the best response.1?? what does that over circle mean??

freckles
 one year ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=limit%28%28%7Cz%2Bh%7C%7Cz%7C%29%2Fh%2Ch%3D0%29

freckles
 one year ago
Best ResponseYou've already chosen the best response.1oh but I think h there is meaning change in real number only

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4nvm, i was thinking stupid something like staying on a circle around 3+i2 for finding f'(3+i2) dw:1436699672546:dw

ytrewqmiswi
 one year ago
Best ResponseYou've already chosen the best response.1i have i more reason which proves that u can't do differentiation with complex numbers

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Yes, @freckles if we are approaching \(3+i2\) along real axis, then \(\Delta y = 0\) : \[\begin{align}\dfrac{dz}{dz}\Bigg_{\Delta y=0} &= \lim\limits_{(\Delta x\to 0, ~\Delta y = 0)} \dfrac{(x+\Delta x)+i(y+0)  x+iy}{\Delta x + i0}\\~\\ &= \lim\limits_{(\Delta x\to 0, ~\Delta y = 0)} \dfrac{(x+\Delta x)+i(y+0)  x+iy}{\Delta x } \\~\\ &= \dfrac{x}{\sqrt{x^2+y^2}}\\~\\ &=\dfrac{x}{z} \end{align}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4If we are approaching \(3+i2\) along imaginary axis, then \(\Delta x = 0\) : \[\begin{align}\dfrac{dz}{dz}\Bigg_{\Delta x=0} &= \lim\limits_{(\Delta x=0, ~\Delta y \to 0)} \dfrac{(x+ 0 )+i(y+\Delta y)  x+iy}{0 + i\Delta y}\\~\\ &= \frac{1}{i}\lim\limits_{(\Delta x=0, ~\Delta y \to 0)} \dfrac{x+i(y+\Delta y)  x+iy}{\Delta y } \\~\\ &= \dfrac{iy}{\sqrt{x^2+y^2}}\\~\\ &=\dfrac{iy}{z} \end{align}\]

ytrewqmiswi
 one year ago
Best ResponseYou've already chosen the best response.1reason  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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4That looks interesting, are you saying \(z^2 = a^2+b^2\) having multiple solutions is somehow bad for differentiability of \(f(z) = z\) ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 have you ever head of something called cauchy riemann equation?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1like 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1since you know z is complex

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[f(x+iy) = u(x,y) + iv(x,y)\] \(f\) is differentiable if and only if \[u_x = v_y~~~\land~~~u_y=v_x\]

ytrewqmiswi
 one year ago
Best ResponseYou've already chosen the best response.1Holomorphic Function?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4that 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1i think you are right

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4If it does work, then we get a pretty interesting result : The derivative of any real valued complex function, if it exists, is \(0\).

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Because 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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