### Author Topic: Hint for Question 20 in Section 1.2  (Read 2457 times)

#### mikerah

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##### Hint for Question 20 in Section 1.2
« on: September 23, 2018, 10:28:15 AM »
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
« Last Edit: September 23, 2018, 12:23:26 PM by Victor Ivrii »

#### Victor Ivrii

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##### Re: Hint for Question 20 in Section 1.2
« Reply #1 on: September 23, 2018, 12:28:22 PM »
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

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##### Re: Hint for Question 20 in Section 1.2
« Reply #2 on: September 24, 2018, 03:39:43 PM »
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.
« Last Edit: September 25, 2018, 01:00:41 PM by Min Gyu Woo »

#### Ye Jin

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##### Re: Hint for Question 20 in Section 1.2
« Reply #3 on: September 24, 2018, 08:43:09 PM »
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.
« Last Edit: September 24, 2018, 08:54:37 PM by Ye Jin »

#### Min Gyu Woo

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##### Re: Hint for Question 20 in Section 1.2
« Reply #4 on: September 27, 2018, 09:17:48 PM »
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  $z_2\mapsto z_2-z_1$.