### Author Topic: Past quiz 1  (Read 962 times)

#### Zhekai Pang

• Newbie
• • Posts: 3
• Karma: 0 ##### Past quiz 1
« on: September 19, 2020, 11:17:42 PM »
http://forum.math.toronto.edu/index.php?topic=1278.0

Why is it a whole line? I also think Re$(z) \geq 0$ is required because the original equation is Re$(z) = | \cdot|$.
« Last Edit: September 19, 2020, 11:43:07 PM by Zhekai Pang »

#### RunboZhang

• Sr. Member
•    • Posts: 51
• Karma: 0 ##### Re: Past quiz 1
« Reply #1 on: September 20, 2020, 09:59:52 AM »
I think it is explained by the first reply in that link. |x+yi-i|=Rez is equivalent to x^2 + (y-1)^2=x^2, subtract both sides by x^2 and this equation would be irrelevant to x. So the only restriction is on y and we have to let y=1. Also, if put it in geometry, it does not matter whether Rez has a positive sign or a negative sign since |x+yi-i| means that the distance between (x,y) and (0,1) is fixed and equal to Rez=x, and it has two corresponding points, (x,1) and (-x,1).

#### Zhekai Pang

• Newbie
• • Posts: 3
• Karma: 0 ##### Re: Past quiz 1
« Reply #2 on: September 20, 2020, 12:09:42 PM »
I disagree. By definition of norm, it is nonnegative. Thus, in order for Re$(z) = |\cdot|$, it has to be nonnegative. You cannot make the first equivalence. For example, $|z| = -3$ is an empty set, although $x^2+y^2=9$ is a circle.

#### Victor Ivrii ##### Re: Past quiz 1
« Reply #3 on: September 20, 2020, 03:02:23 PM »
half-line