If $u(z),$ where $Z_t=W^1_t-iW^2_t$, is a complex anlyt. fx, show $du(Z_t)=u'(Z_t)dZ_t$

Question

If $u(z)$ is a complex analytical function, where $Z_t=W^1_t-iW^2_t$ is a complex Wiener process, show $du(Z_t)=u'(Z_t)dZ_t$.

Answer 1

To make your tutor happy two things. Firstly, $u$ has no time dependence so the derivative of $u$ wrt $t$ is zero. Also with $f(z)=a(z)+ib(z)$ you have \begin{equation} \frac{1}{2}\frac{\partial^{2}f}{\partial z^{2}}d(Z_{t})^{2}+\frac{1}{2}\frac{\partial^{2}f}{\partial \bar{z}^{2}}d(\bar{Z_{t}})^{2}= \frac{1}{2}\left( \frac{\partial^{2}a}{\partial (x^{1})^{2}} + \frac{\partial^{2}a}{\partial (x^{2})^{2}} +i\left(\frac{\partial^{2}b}{\partial (x^{1})^{2}}+\frac{\partial^{2}b}{\partial (x^{2})^{2}} \right) \right)dt. \end{equation} Now use the Cauchy-Riemann equations to show this is zero and you are done.

Answer 2

Ok Solved it:

Using Ito's formula for multidimensional function I know $dZ_t=dW^1_t + idW^2_t.$ Also Ito's formula for Complex function is:

$df(Z_t)=\frac{\partial f}{\partial z}d(Z_t) + \frac{\partial f}{\partial \bar{z}}d(\bar{Z}_t) + \frac{1}{2}\frac{\partial^2 f}{{\partial z}^2}d(Z_t)d(Z_t) + \frac{1}{2}\frac{\partial^2 f}{{\partial \bar{z}}^2}d(\bar{Z}_t)d(\bar{Z}_t) + \frac{\partial^2 f}{\partial z \partial \bar{z}}d(Z_t)d(\bar{Z}_t)$.

Set f(z)=u(z). Since u(z) is analytic, $\frac{\partial u}{\partial \bar{z}} =0$. Also $dZ_t dZ_t = 0$ using the facts that $dW^i_tdW^i_t=dt$ and $dW^i_tdW^j_t=0$. Therefore

$du(Z_t)=\frac{\partial u}{\partial z}d(Z_t)$ as needed :)