This question is from the paper 'A new proof of Spitzer's result on the winding of two dimensional Brownian motion' by R Durrett, 1982.
Let $D_t = A_t + iB_t$ be a complex Brownian motion. Then $\int \vert e^{iD_s} \vert ^2 ds = \int e^{2A_s} ds$.
But why? I think by Eulers equation should follow: $\vert e^{iA_s + i^2B_s} \vert ^2 = \vert e^{iA_s}e^{-B_s} \vert ^2= \vert e^{-B_s} \vert ^2 = e^{-2B_s}$
I'd appreciate the help! Thanks :))
Let's clarify the misunderstanding here. Your reasoning is almost correct, but there's a small mistake in the application of Euler's equation. Let's go through it step by step:
Starting from Euler's equation: $$ \left| e^{i A_s + i B_s} \right|^2 = \left| e^{i A_s} e^{i B_s} \right|^2. $$
Now, we have: $$ \left| e^{i A_s} e^{i B_s} \right|^2 = \left| e^{i A_s} \right|^2 \left| e^{i B_s} \right|^2. $$
Using Euler's equation for each term separately: $$ \left| e^{i A_s} \right|^2 = \left| \cos(A_s) + i \sin(A_s) \right|^2 = \cos^2(A_s) + \sin^2(A_s) = 1, $$ and $$ \left| e^{i B_s} \right|^2 = \left| \cos(B_s) + i \sin(B_s) \right|^2 = \cos^2(B_s) + \sin^2(B_s) = 1. $$
Hence, $$ \left| e^{i A_s + i B_s} \right|^2 = \left| e^{i A_s} \right|^2 \left| e^{i B_s} \right|^2 = 1 \cdot 1 = 1. $$
This means that $\left| e^{i D_s} \right|^2 = 1$, which is not the same as $\left| e^{i D_s} \right|^2 = e^{2 A_s}$.
The correct relation should follow from the fact that $D_t = A_t + iB_t$ is a complex Brownian motion, where both $A_t$ and $B_t$ are real-valued Brownian motions. Then, using the property of the exponential function: $$ \left| e^{i D_s} \right|^2 = e^{2 \text{Re}(iD_s)} = e^{2 A_s}. $$
Therefore, $\left| e^{i D_s} \right|^2 = \left| e^{i D_s} \right|^2 = e^{2 A_s}$ as desired.