MAT334--2020F > Chapter 1

Past quiz 1

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**Zhekai Pang**:

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$(z) = | \cdot|$.

**RunboZhang**:

Reply to Zhekai:

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**:

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**:

half-line

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