What does it mean for a point to be on the segment $[1,i]$?

Question

I came up on a problem where I have to prove that there does not exist a point $a\in\mathbb{C}$ on the segment $[1,i]$ such that $i+1=3a^2$.

I am struggling to understand what is meant by $a$ being on the segment $[1,i]$. Does it mean that $a$ is on the segment from $(1,0)$ to $(0,i)$ on the complex plane? Or does it means something different?

Answer 1

The line segment from $(1,0)$ to $(0,i)$ is the set of points $(t, i(1-t)), 0 \leq t \leq 1$. If this satisfies the given equation you get $1=2t-1$ and $1=2t(1-t)$. The first equation gives $t=2/3$ so the second equation cannot hold.

Answer 2

No; it means that it belongs to $\{\lambda i+(1-\lambda)\mid\lambda\in[0,1]\}$. This set is precisely the line segmen joining $1$ to $i$. Note that $\lambda i+(1-\lambda)$ is equal to $1$ when $\lambda=0$ and it is equal to $i$ when $\lambda=1$.

Answer 3

As others have noted, the "segment $[1,i]$" most likely means the line segment in the complex plane connecting $1$ on the real axis to $i$ on the imaginary axis. Note that the nearest this gets to the origin is at its midpoint, $(1+i)/2$, for which the modulus is $|(1+i)/2|=1/\sqrt2$. Consequently, if $a$ is any point on the segment, we see that

$$3|a|^2\ge3/2\gt\sqrt2=|1+i|$$

and thus it's not possible to have $3a^2=1+i$.