Gaussian plane 1

[rsoldo]

Let $S=\left\{z\in C: Re((1+i)z)+Im\left(\frac{z}{2-i}\right)=1\right\}$ and $T=\left\{\frac{1}{z}:z\in S\right\}$. Determine and draw in Gaussian plane these sets.

[мат-ламер]

(Оффтоп)

А это точно олимпиадная задача? Простовато как-то выглядит.

[Dendr]

(Оффтоп)

Куда уж проще, понимания комплексных чисел более чем достаточно

А чтобы тема не пропадала без ответа...



NB. На рисунке дополнительно показан ход графического решения для $T$.

[rsoldo]

How to calculate the equation of a circle?

[mihiv]

Уравнение окружности: $(x-a)^2+(y-b)^2=r^2$, уравнение прямой: $y=2x-\frac 53$.

Если $y=0$, то $x=\frac 56, z=\frac 56, \dfrac 1z=1.2, a=\frac {1.2}2=0.6$, если $x=0$,то $y=-\frac 53, z=-i\frac 53, \dfrac 1z=0.6i, b=\dfrac {0.6}2=0.3, r^2=(0-0.6)^2+(0-0.3)^2=\dfrac 9{20}$

[rsoldo]

Nice. But how do we know that set of points is a circle? We need to prove that.

[zykov]

rsoldo в сообщении #1542326 писал(а):

But how do we know that set of points is a circle?

Pythagorean theorem

[zykov]

If the question is about why $\frac{1}{z}$ transforms a circle to a line and back, then it is like this.
Circle equation (center at $z_0$, passing through $0$):

$$|z-z_0|=|z_0|$$

Substitute $\frac{1}{z}$ for $z$ (note $z \neq 0$):

$$\left(\frac{1}{z}-z_0 \right) \overline{\left(\frac{1}{z}-z_0 \right)}=z_0 \overline{z_0}$$

$$\frac{1}{z \overline z}-\frac{z_0}{\overline z} -\frac{\overline{z_0}}{z}+z_0 \overline{z_0}=z_0 \overline{z_0}$$

$$1 = z z_0 +\overline{z z_0}$$

Hence $\operatorname{Re} (z z_0) = \frac12$ and therefore $\operatorname{Re} z \operatorname{Re} z_0 - \operatorname{Im} z \operatorname{Im} z_0 = \frac12$, means $x x_0 - y y_0 = \frac12$ - equation of a line.

[rsoldo]

$z=x+y\cdot i$

$(1+i)(x+yi)=x+yi+xi-y=(x-y)+(x+y)i$

$\frac{x+yi}{2-i}\cdot \frac{2+i}{2+i}=\frac{2x-y}{5}+\frac{x+2y}{5}\cdot i$
From the conditions of the problem we get the equation:
$6x-3y-5=0$

Now, $w=\frac{1}{z}=u+vi, u,v\in \mathbb{R}, w\neq 0$
we have
$z=\frac{1}{w}=\frac{1}{u+vi}\cdot \frac{u-vi}{u-vi}=\frac{u}{u^2+v^2}-\frac{v}{u^2+v^2}i$
and
$x=\frac{u}{u^2+v^2} , y=\frac{-v}{u^2+v^2}$
From the equation of line:
$\implies \frac{6u}{u^2+v^2}+\frac{3v}{u^2+v^2}=5$

$\implies 5u^2-6u+5v^2-3v=0$

$\implies u^2-\frac{6}{5}u+\frac{9}{25}+v^2-\frac{3}{5}v+\frac{9}{100}=\frac{9}{25}+\frac{9}{100}$

$\implies \left(u-\frac{3}{5}\right)^2+\left(v-\frac{3}{10}\right)^2=\frac{9}{20}$

We have a picture above!