Formula for the propagation of complex errors

[accdd]

If I have 2 measurements $x = (3.0 ± 0.1), y = (-2.0 ± 0.1)$ and want to calculate how the error propagates when calculating a function from those values this formula should be used: $f(x, y) = f(x, y) ± \sqrt {(\frac{\partial f}{\partial x}*\Delta x)^2+(\frac{\partial f}{\partial y}*\Delta y)^2}$
What is the formula for calculating error propagation if x and y are complex ($x = (3 ± 0.1) + (9.5 ± 0.4)ⅈ, y = (2 ± 0.1) - (5 ± 0.4)ⅈ$)?

 

[Svein]

Change to polar coordinates.

 

[accdd]

Can you give me an example. Suppose the function is: $f(x, y) = x + y^2$
In the non-complex case, with the data given in the previous post, I would proceed as follows:
$f(x, y) = x + y^2 = (3+(-2)^2) \pm \sqrt{0.1^2+(2*(-2) *0.1)^2}= 7\pm \sqrt{0.01+0.16}=7\pm0.41$
How can I change to polar coordinates to get the result in case of complex numbers?
The result should be: $(-18.0 ± 4.0) - (10.5 ± 1.9)ⅈ$ (Measurement jl)

 

[Svein]

$x+iy\equiv Re^{i\theta}$where $R=\sqrt{x^{2}+y^{2}}$ and $\theta =\arctan(\frac{y}{x})$. This is the easiest representation for complex multiplication (you multiply the argumets and add the angles). Complex addition is easiest in the cartesian notation.

 

[accdd]

I got stuck, not able to get the result. Can someone show me how to do it please?

 

[Stephen Tashi]

Change to polar coordinates.

After that, you'd have to know the variance of the amplitude as a function of the variances of the real and imaginary components. If they had the same variance, I think you'd have a Rayleigh distribution. I'm not sure how that generalizes to the case of unequal variances. Does that distribution have a special name?