Show that $\mathbb{C}\setminus \mathbb{R}^{\le 0}$ is simply connected [duplicate]

I need to show that for $\Bbb C/(-\infty ,0]$ that if I have arbitrary different pathes $\gamma , \eta :[0,1] \to\Bbb C/(-\infty ,0]$ such that $\gamma(0)=\eta(0)=\alpha $ and $\gamma(1)=\eta(1)=\beta $ I have continuous homotopy $H:[0,1]^2 \to \Bbb C/(-\infty ,0]$ such that: $H(0,t) = \gamma (t),H(1,t) = \eta (t),H(s,0) = \alpha,H(s,1) = \beta$ unfortunatily my space isn`t a convex,so how do I define a proper Homotopy and proving it's continuous?

The simplest way is probably to use that your domain is starshaped with respect to 1. This makes it possible to shrink any closed curve in the domain to the point 1.

Consider the complex plane using polar coordinate system. Every points in $\mathbb{C}\setminus(-\inf,0]$ can be written as $(\rho,\theta)$, $\rho\in(0,+\inf)$, $\theta\in(-\pi,\pi)$. The function $f\colon\mathbb{C}\setminus(-\inf,0]\to\{z\in\mathbb{C}, Im(z)>0\}$ defined as $f(\rho,\theta)=(\rho,\frac{\theta}{2})$ is an homeomorphism, and Imf is a convess space, so the domain is simply connected.