Q1: TUT 0101 and TUT 0102

[Victor Ivrii]

$\renewcommand{\Re}{\operatorname{Re}}
\renewcommand{\Im}{\operatorname{Im}}$
Describe the locus of points $z$ satisfying the given equation.
\begin{equation*}
| z - i| = \Re z.
\end{equation*}

[Ye Jin]

|x+yi-i|=Rez
|x+(y-1)i|=x
Square Root [x^2+(y-1)^2]=x
x^2 + (y-1)^2=x^2
(y-1)^2=0
So, y=1

[Vedant Shah]

Let $z=x+iy$
Then $Re(z) = x$ and $|z-i| = |x+i(y-1)|$
Thus:
$Re(z) = |z-i|$
$x = |x+i(y-1)|$
$x = \sqrt{x^2 + (y-1)^2}$
$x^2 = x^2 +(y-1)^2, x \ge 0$
$(y-1)^2 = 0, x \ge 0$
$y=1, x \ge 0$

In complex terms:
$ y = 1 \iff Im(z) = 1$ and $ x \ge 0 \iff Re(z) \ge 0$

Thus the equation of the line in complex terms is:
$Im(z) = 1, Re(z) \ge 0$

This is the horizontal half line extending from $z=i$ rightward.

[Victor Ivrii]

Vedant,
it is a half line. Also, no need to post after solution is posted