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

- ganeshie8

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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

Yes \(z = a+ib\)
then \(|z| = \sqrt{a^2+b^2}\)

- 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 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

\[\dfrac{d|z|}{dz} = \lim\limits_{\Delta z\to 0} \dfrac{|z+\Delta z| - |z|}{\Delta z}\]
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

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

If \(z=a+ib\),
then \(a\) and \(b\) change as \(z\) changes, so they are not really constants with respect to \(z\) here

- UsukiDoll

oh my O_O! so a and b is connected to z then

- 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

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

@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

let me know if tht above is like really really wrong or whatever

- 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

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

For the limit of a function of two variables to exist,
the limit must exist in all directions and be the same

- 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

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

- ganeshie8

|dw:1436697803754:dw|

- ganeshie8

\(f'(3+i2)\) in each direction need not be same, so the derivative doesn't exist

- ganeshie8

Wow! how did you get that
wil try to figure it out... have good sleep :)

- sdfgsdfgs

@ganeshie8 think about delta r / delta z n what u just posted about the "direction" (which is theta...

- 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

\[\frac{d \mid z \mid}{dz} = \cos(\theta) +isin(\theta), z =(r, \theta) \]

- ganeshie8

\[\large \frac{d|z|}{dz} = e^{i\arg(z)}\]
?
this looks interesting

- 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

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

- freckles

http://mathworld.wolfram.com/AbsoluteValue.html

- freckles

d|z|/dz doesn't exist according to this
giving an explanation about direction as @ganeshie8 was saying

- 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

I'm going to work the derivatives in \(x=0\), \(y=0\) and \(y=mx\) directions and see if they are different

- 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

if you know anyways (I know you are the wolfram master)

- 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

oh

- ganeshie8

As expected it says there are no solutions

- freckles

I completely misread that

- 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

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

\[\dfrac{d|z|}{dz} \Bigg|_{\text{over circle}} = 0\]
?

- ytrewqmiswi

?? what does that over circle mean??

- ytrewqmiswi

integration?

- freckles

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

- freckles

oh but I think h there is meaning change in real number only

- ganeshie8

nvm, i was thinking stupid something like staying on a circle around 3+i2 for finding f'(3+i2)
|dw:1436699672546:dw|

- ytrewqmiswi

i have i more reason which proves that u can't do differentiation with complex numbers

- 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

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

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

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

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

yes

- freckles

@ganeshie8 have you ever head of something called cauchy riemann equation?

- freckles

heard*

- 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

since you know z is complex

- 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

Holomorphic Function?

- 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

i think you are right

- 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

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.