ganeshie8
  • ganeshie8
\[\large \dfrac{d|z|}{dz} = ?\] \(z\) is a complex number
Mathematics
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katieb
  • katieb
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UsukiDoll
  • UsukiDoll
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 :/
ganeshie8
  • ganeshie8
Yes \(z = a+ib\) then \(|z| = \sqrt{a^2+b^2}\)
UsukiDoll
  • UsukiDoll
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.

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UsukiDoll
  • UsukiDoll
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?
ganeshie8
  • ganeshie8
\[\dfrac{d|z|}{dz} = \lim\limits_{\Delta z\to 0} \dfrac{|z+\Delta z| - |z|}{\Delta z}\] Let \(z=x+iy\), then the above is is same as \[\dfrac{d|z|}{dz} = \lim\limits_{(\Delta x\to 0, ~\Delta y\to 0)} \dfrac{|(x+\Delta x)+i(y+\Delta y)| - |x+iy|}{\Delta x + i\Delta y}\]
sparrow2
  • sparrow2
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 :)
ganeshie8
  • ganeshie8
If \(z=a+ib\), then \(a\) and \(b\) change as \(z\) changes, so they are not really constants with respect to \(z\) here
UsukiDoll
  • UsukiDoll
oh my O_O! so a and b is connected to z then
sdfgsdfgs
  • sdfgsdfgs
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.
freckles
  • freckles
what if we think of a and b as real-valued functions of x then z is too \[|z|(x)=\sqrt{[a(x)]^2+[b(x)]^2} \\ |z|'(x)=\frac{1}{2 \sqrt{a^2(x)+b^2(x)}} (2a(x)a'(x)+2 b(x) b'(x)) \\ |z|'(x)=\frac{aa'+bb'}{\sqrt{a^2+b^2}} \\ \text{ So we have } \\ \frac{d}{dx}|z|=\frac{aa'+bb'}{\sqrt{a^2+b^2}} \\ d|z|=\frac{aa'+bb'}{\sqrt{a^2+b^2}}dx \\ \frac{d|z|}{dz}=\frac{aa'+bb'}{\sqrt{a^2+b^2}} \frac{dx}{dz} \\ z(x)=a(x)+b(x) \cdot i \\ \frac{dz}{dx}=a'+b' \cdot i \\ \frac{d|z|}{dz}=\frac{aa'+bb'}{\sqrt{a^2+b^2}} \frac{1}{a'+b' i}\]
ganeshie8
  • ganeshie8
@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
  • freckles
let me know if tht above is like really really wrong or whatever
sdfgsdfgs
  • sdfgsdfgs
@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
  • ganeshie8
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 \(z\)
ganeshie8
  • ganeshie8
For the limit of a function of two variables to exist, the limit must exist in all directions and be the same
freckles
  • freckles
so we are saying \[\frac{d|z|}{dz}=\frac{aa'+bb'}{\sqrt{a^2+b^2}} \frac{1}{a'+b' i} \text{ doesn't exist } ? \\ \text{ so } z=2x -5i \\ \text{ so } \frac{d|z|}{dz} \text{ doesn't exist } \\ \text{ or is \it } \frac{4x}{\sqrt{4x^2+25}} \frac{1}{2-0i} =\frac{2x}{\sqrt{4x^2+25} }\]
ganeshie8
  • ganeshie8
I think we're saying \(\dfrac{d|z|}{dz}\) exists in a specific direction but it is not the same in all directions
ganeshie8
  • ganeshie8
|dw:1436697803754:dw|
ganeshie8
  • ganeshie8
\(f'(3+i2)\) in each direction need not be same, so the derivative doesn't exist
ganeshie8
  • ganeshie8
Wow! how did you get that wil try to figure it out... have good sleep :)
sdfgsdfgs
  • sdfgsdfgs
@ganeshie8 think about delta r / delta z n what u just posted about the "direction" (which is theta...
sdfgsdfgs
  • sdfgsdfgs
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!
UsukiDoll
  • UsukiDoll
\[\frac{d \mid z \mid}{dz} = \cos(\theta) +isin(\theta), z =(r, \theta) \]
ganeshie8
  • ganeshie8
\[\large \frac{d|z|}{dz} = e^{i\arg(z)}\] ? this looks interesting
ytrewqmiswi
  • ytrewqmiswi
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.
ganeshie8
  • ganeshie8
In derivatives/limits we're dealing with absolute values of complex numbers, which are real. So it should be fine ?
freckles
  • freckles
http://mathworld.wolfram.com/AbsoluteValue.html
freckles
  • freckles
d|z|/dz doesn't exist according to this giving an explanation about direction as @ganeshie8 was saying
sdfgsdfgs
  • sdfgsdfgs
@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!
ganeshie8
  • ganeshie8
I'm going to work the derivatives in \(x=0\), \(y=0\) and \(y=mx\) directions and see if they are different
freckles
  • freckles
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.
freckles
  • freckles
if you know anyways (I know you are the wolfram master)
ganeshie8
  • ganeshie8
Looks there is an issue with the input interpretation, wolf is treating the input as \[\text{solve}~~\dfrac{d|z|}{dz} = 0~~\text{over complex numbers}\]
freckles
  • freckles
oh
ganeshie8
  • ganeshie8
As expected it says there are no solutions
freckles
  • freckles
I completely misread that
ytrewqmiswi
  • ytrewqmiswi
@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
  • freckles
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
ganeshie8
  • ganeshie8
\[\dfrac{d|z|}{dz} \Bigg|_{\text{over circle}} = 0\] ?
ytrewqmiswi
  • ytrewqmiswi
?? what does that over circle mean??
ytrewqmiswi
  • ytrewqmiswi
integration?
freckles
  • freckles
http://www.wolframalpha.com/input/?i=limit%28%28%7Cz%2Bh%7C-%7Cz%7C%29%2Fh%2Ch%3D0%29
freckles
  • freckles
oh but I think h there is meaning change in real number only
ganeshie8
  • ganeshie8
nvm, i was thinking stupid something like staying on a circle around 3+i2 for finding f'(3+i2) |dw:1436699672546:dw|
ytrewqmiswi
  • ytrewqmiswi
i have i more reason which proves that u can't do differentiation with complex numbers
freckles
  • freckles
http://www.wolframalpha.com/input/?i=limit%28%28%7C%28a%2Bbi%29%2B%28z%29%7C-%7Ca%2Bbi%7C%29%2F%28z%29%2Cz%3D0%29
ganeshie8
  • ganeshie8
Yes, @freckles if we are approaching \(3+i2\) along real axis, then \(\Delta y = 0\) : \[\begin{align}\dfrac{d|z|}{dz}\Bigg|_{\Delta y=0} &= \lim\limits_{(\Delta x\to 0, ~\Delta y = 0)} \dfrac{|(x+\Delta x)+i(y+0)| - |x+iy|}{\Delta x + i0}\\~\\ &= \lim\limits_{(\Delta x\to 0, ~\Delta y = 0)} \dfrac{|(x+\Delta x)+i(y+0)| - |x+iy|}{\Delta x } \\~\\ &= \dfrac{x}{\sqrt{x^2+y^2}}\\~\\ &=\dfrac{x}{|z|} \end{align}\]
ganeshie8
  • ganeshie8
If we are approaching \(3+i2\) along imaginary axis, then \(\Delta x = 0\) : \[\begin{align}\dfrac{d|z|}{dz}\Bigg|_{\Delta x=0} &= \lim\limits_{(\Delta x=0, ~\Delta y \to 0)} \dfrac{|(x+ 0 )+i(y+\Delta y)| - |x+iy|}{0 + i\Delta y}\\~\\ &= \frac{1}{i}\lim\limits_{(\Delta x=0, ~\Delta y \to 0)} \dfrac{|x+i(y+\Delta y)| - |x+iy|}{\Delta y } \\~\\ &= \dfrac{-iy}{\sqrt{x^2+y^2}}\\~\\ &=-\dfrac{iy}{|z|} \end{align}\]
ytrewqmiswi
  • ytrewqmiswi
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
ganeshie8
  • ganeshie8
That looks interesting, are you saying \(|z|^2 = a^2+b^2\) having multiple solutions is somehow bad for differentiability of \(f(z) = |z|\) ?
ytrewqmiswi
  • ytrewqmiswi
yes
freckles
  • freckles
@ganeshie8 have you ever head of something called cauchy riemann equation?
freckles
  • freckles
heard*
freckles
  • freckles
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
freckles
  • freckles
since you know z is complex
ganeshie8
  • ganeshie8
\[f(x+iy) = u(x,y) + iv(x,y)\] \(f\) is differentiable if and only if \[u_x = v_y~~~\land~~~u_y=-v_x\]
ytrewqmiswi
  • ytrewqmiswi
Holomorphic Function?
ganeshie8
  • ganeshie8
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
freckles
  • freckles
i think you are right
ganeshie8
  • ganeshie8
If it does work, then we get a pretty interesting result : The derivative of any real valued complex function, if it exists, is \(0\).
ganeshie8
  • ganeshie8
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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