# Formula for the propagation of complex errors

• I
• accdd
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##)?

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)

$x+iy\equiv Re^{i\theta}$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?

Svein said:
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?

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