 Author Topic: Q1: TUT 0101 and TUT 0102  (Read 1135 times)

Victor Ivrii Q1: TUT 0101 and TUT 0102
« on: September 28, 2018, 04:09:13 PM »
$\renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}}$
Describe the locus of points $z$ satisfying the given equation.
\begin{equation*}
| z - i| = \Re z.
\end{equation*}

Ye Jin

• Full Member
•   • Posts: 17
• Karma: 9 Re: Q1: TUT 0101 and TUT 0102
« Reply #1 on: September 28, 2018, 04:36:18 PM »
|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

• Jr. Member
•  • Posts: 13
• Karma: 8 Re: Q1: TUT 0101 and TUT 0102
« Reply #2 on: September 28, 2018, 06:01:05 PM »
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 Re: Q1: TUT 0101 and TUT 0102
« Reply #3 on: September 29, 2018, 03:44:43 PM »
Vedant,
it is a whole line. Also, no need to post after solution is posted