Question

\[\large \dfrac{d|z|}{dz} = ?\]

\(z\) is a complex number

Answer 1

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 :/

Answer 2

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

Answer 3

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.

Answer 4

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?

Answer 5

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

Let \(z=x+iy\), then the above is is same as

\[\dfrac{d|z|}{dz} = \lim\limits_{(\Delta x\to 0, ~\Delta y\to 0)} \dfrac{|(x+\Delta x)+i(y+\Delta y)| - |x+iy|}{\Delta x + i\Delta y}\]

Answer 6

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 :)

Answer 7

If \(z=a+ib\),

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

Answer 8

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

Answer 9

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.

Answer 10

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

Answer 11

@sdfgsdfgs why am i not allowed to replace \(\Delta z\) by \(\Delta r\) and get
\[\frac{ |z+ \Delta z| - |z| }{ \Delta z } = \frac{ r + \Delta r - r }{ \Delta r } = \frac{ \Delta r }{ \Delta r} = 1\]

?

Answer 12

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

Answer 13

@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.

Answer 14

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 \(z\)

Answer 15

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

Answer 16

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

Answer 17

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