Hint for Question 20 in Section 1.2

[mikerah]

Question 20 in section 1.2 is quite intuitive to see that it is true. However, I am having difficulties starting to prove what agrees with my intuition.

Currently, I am trying to use the argument to show that $tz_1+(1-t)z_2$ is indeed the line joining $z_1$ and $z_2$.

Does anyone have any hints as to how to start proving Question 20?

Thank you

[Victor Ivrii]

Select an origin at $z_1$. Describe what you get for

* $t=0$, $t=1$
* $0<t<1$, in particular $t=\frac{1}{2}$
* $t<0$, in particular $t=-1$
* $t>0$, in particular $t=2$

You MathJax for math snippets,

[Min Gyu Woo]

So if we set the origin to be $z_1$,

Is it enough to show that the $tz_1 + (1-t)z_2$ follows the equation of the line

$Re(-(t-1)(1+i)z_2 + 0)=0$ which simplifies to $y=x$ ?  How that?! V.I.

EDIT:

$Re(-(t-1)(1+i)z_2) =0$

If $a=(1-t)(1+i)=(1-t)+i(1-t) \in \mathbb{C}$

We can simplify to

$Re(az_2 + 0) = (1-t)x -(1-t)y + Re(0) = 0$

Then we get

$(1-t)x=(1-t)y$

$x=y$

NOTE: When $t=1$ we get $z_1 = 0$ which we already have if we centered $z_1$ at the origin.

[Ye Jin]

Could you explain more about the last two situations？I do not know how to deal with it when either of t or 1-t becomes negative especially when t=2 or -1.

[Min Gyu Woo]

I thought of it this way. Hope this helps.

[Victor Ivrii]

It is a way more simple. Assume first that $z_1=0$. What we get then?

Reduce the general case to the previous one by  moving a coordinate system so $z_1\mapsto 0$: $z\mapsto z-z_1$, then  $z_2\mapsto z_2-z_1$.