$\gamma'_z(t)=(ie^{it}z_1,ie^{it}z_2)$ in stereographic coordinates is non-vanishing for all $t$

Question

I have to solve this one:

Consider $S^3$ as a subset of $\mathbb{C}^2$. For all $(z_1,z_2)\in S^3$, consider $\gamma_z:\mathbb{R}\rightarrow S^3$ defined by $\gamma_z(t)=(e^{it}z_1,e^{it}z_2)$.

1. Give the representation in stereographic coordinates of $\gamma_z$ and use it to prove that $\gamma_z$ is a smooth curve.

2. Compute the tangent vector $\gamma'_z(t)$ in stereographic coordinates and prove that is non-vanishing for all $t$.

Solution:

1. $(z_1,z_2)=(a+ib,c+id)\in S^3$, \begin{eqnarray*} \gamma_z(z_1,z_2)&=&\Big(\big(\cos(t)+i\sin(t)\big)\big(a+ib\big),\big(\cos(t)+i\sin(t)\big)\big(c+id\big)\Big)\\ &=& \Big( \underbrace{\big(a\cos(t)-b\sin(t)\big)+i\big(a\sin(t)+b\cos(t)\big)}_{u+iv},\underbrace{\big(c\cos(t)-d\sin(t)\big)+i\big(c\sin(t)+d\cos(t)\big)}_{r+is} \Big)\\ &=& (u+iv,r+is)\cong (u,v,r,s)\in\mathbb{R}^4 \end{eqnarray*}

Then let $\varphi_N$ be the stereographic projection from the north pole $(0,i)\in\mathbb{C}^2$ we have

\begin{eqnarray*} \varphi_N(\gamma_z(z_1,z_2))&=&\frac{1}{1-s}\Big(u,v,r\Big)\in \mathbb{R}^3 \end{eqnarray*}

Since every component is smooth, $\forall t\in\mathbb{R}$, $\gamma_z$ is smooth on $S^3$.

2.

\begin{eqnarray*} \gamma'_z(z_1,z_2)&=&(ie^{it}z_1,ie^{it}z_2)\\ &=& \Big(\underbrace{-\big(a\sin(t)+b\cos(t)\big) +i\big(a\cos(t)-b\sin(t)\big)}_{-v+iu},\underbrace{-\big(c\sin(t)+d\cos(t)\big)+i\big(c\cos(t)-d\sin(t)\big)}_{-s+ir} \Big)\\ &=& (-v+iu,-s+ir)\cong (-v,u,-s,r)\in\mathbb{R}^4 \end{eqnarray*} and with the stereographic projection we get

\begin{eqnarray*} \varphi_N(\gamma_z(z_1,z_2))&=&\frac{1}{1-r}\Big(-v,u,-s\Big)\in \mathbb{R}^3 \end{eqnarray*} This is the tangent vector $\gamma'_z$ in stereographic coordinates.

Now the problem is that I don't know how to check that $\gamma'_z$ is non-vanishing on $S^3$ for all $t\in \mathbb{R}$.

Can you help me?

Thanks a lot