Show that the exponential function maps any line through the origin in $\mathbb{R}i +\mathbb{R}j + \mathbb{R}k$ onto a circle of radius $1$ in $\mathbb{S}^3$.
I know that for any element v $\in\mathbb{R}i +\mathbb{R}j + \mathbb{R}k$, we can write $v = \theta u$, where $u$ is a unit vector in $\mathbb{R}i +\mathbb{R}j + \mathbb{R}k$ and $\theta$ is any real number.
Then $e^{\theta u} = cos\theta + usin\theta\in S^3$ (i.e we can map any pure imaginary quaternion onto $S^3 = SU(2)$ So the exponential function maps $\mathbb{R}i +\mathbb{R}j + \mathbb{R}k$ onto $S^3$
What is a line through $O$ in $\mathbb{R}i +\mathbb{R}j + \mathbb{R}k$? What is a circle of radius $1$ in $\mathbb{S}^3$?
If $u$ is a unit norm element in $\mathbb{R}i+\mathbb{R}j+\mathbb{R}k$ then the elements $1$ and $u$ form an orthonormal basis for the plane spanned by them. Therefore the exponential map you wrote down by definition gives the unit circle in this plane. A line through the origin is simply the line spanned by $u$.