Q) Let $f(B_t)$ be a complex Brownian motion in the complex plane. (This means that $B_t = B_t^1 + iB_t^2$ where $B_t^1$ and $B_t^2$ are independent Brownian motions.) Let $f(z):\mathbb{C} \to \mathbb{C}$ be the entire function $f(z) = e^z$. Show that $f(B_t)$ is a time-change of Brownian motion, and conclude that complex Brownian motion never hits $0$, almost surely.
At a higher level, there is a theorem that says if a continuous local martingale has quadratic variation $\uparrow\infty$, then it is a time change of Brownian motion.
$f(x+iy) = e^x\cos y+i\sin y$. Let $f_1(x+iy) = e^x\cos y$. Then:
$$\frac{\partial f_1}{\partial x} = e^x\cos y, \frac{\partial f_1}{\partial y} = -e^x\sin y, \frac{\partial^2 f_1}{\partial x^2} = e^x\cos y, \frac{\partial^2 f_1}{\partial y^2} = -e^x\cos y\implies \nabla f_1 = 0$$
where $\nabla$ is the Laplacian operator. Now, by Ito's formula it follows that:
$$f(B_t^1,B_t^2) = f_1(0,0)+\int_0^t e^{B_s^1}\cos B_s^2 dB_s^1 - \int_0^t e^{B_s^1}\sin B_s^2 dB_s^2 + \text{ terms related to }f_2$$
Thus the quadratic variation of the two integrals seen above is:
$$\int_0^t e^{2B_s^1}\cos^2 B_s^2 ds + \int_0^t e^{2B_s^1}\sin^2 B_s^2 ds = \int_0^t e^{2B_s^1} ds$$
1) So, how do I show that $\int_0^t e^{2B_s} ds \uparrow \infty$ as $t \uparrow \infty$?
2) Also, how does $f(B_t)$ is a time-change of Brownian motion imply that complex Brownian motion never hits $0$?
For the first question, I will outline a proof of the fact that if $W_s$ is a real-valued Brownian motion then $C = \int_0^\infty e^{2W_s} ds = \infty$. This is sufficient since $\int_0^t e^{2W_s}ds$ is increasing in $t$.
The most simple first guess for how to do this is to bound $C \geq \int_0^\infty 1_{\{W_s \geq 0\}} ds$ and try to show that there almost surely exists an infinite sequence of times $t_n > 1$ such that $W(s) \geq 0$ for all $s \in [t_n - 1, t_n]$ (note that $t_n$ is a random variable here).
One can check that $T = \inf\{t > 1: W_u \geq 0 \text{ for all } u \in [t-1, t]\}$ is a stopping time for the Brownian filtration and so to see the existence of the sequence $t_n$ it suffices to show that $\mathbb{P}(T < \infty) = 1$ and apply the usual recursive argument using the strong Markov property.
Let $H_n = \inf\{t > 0: W_t = n\}$. Then, by the strong Markov property and symmetry, $$\mathbb{P}(W_t < 0 \text{ for some } t \in [H_n, H_n + 1]) = \mathbb{P}( H_n \leq 1).$$ By the reflection principle, $\mathbb{P}(H_n \leq 1) = 2 \mathbb{P}(W_1 \geq n) \to 0$ as $n \to \infty$ since $W_1$ is a standard Gaussian.
In particular, we can deduce that $$\mathbb{P}(\forall n, W_t < 0 \text{ for some } t \in [H_n, H_n + 1]) = 0$$ which in turn implies that $$\mathbb{P}(T < \infty) \geq \mathbb{P}( \exists n, W_t \geq 0 \text{ for all } t \in [H_n,H_n + 1]) = 1.$$
For your second question, if $f(B_s)$ is a time-change of the Brownian motion $\tilde{B}_s$ then the events $A = \{\exists s: f(B_s) = 0\}$ and $B = \{\exists u: \tilde{B}_u = 0\}$ are equal. Since $f(B_s) = e^{B_s}$, you know that $\mathbb{P}(A) = 0$ which implies that $\mathbb{P}(B) = 0$ which is the desired result (for the Brownian motion $\tilde{B}$).