giải thích chi tiết $Ay=0$,N.

Question

Lại Thị Kim Liên · Accepted Answer

Gọi số phức z có dạng $\displaystyle z=a+bi\Longrightarrow \overline{z} =a-bi$
Ta có:
$\displaystyle \begin{array}{{>{\displaystyle}l}}
|z-\overline{z} +1-i|=2\
\Longrightarrow |a+bi-a+bi+1-i|=2\
\Longrightarrow |( 2b-1) i+1|=4\
\Longrightarrow \sqrt{( 2b-1)^{2} +1} =4\
\Longrightarrow ( 2b-1)^{2} +1=16\
\Longrightarrow ( 2b-1)^{2} =15\
\Longrightarrow \left[ \begin{array}{l l}
2b-1=\sqrt{15} & \
2b-1=-\sqrt{15} &
\end{array} ight.\
\Longrightarrow b=\frac{1\pm \sqrt{15}}{2}
\end{array}$

Timi · Answer

To solve the given problem, we can follow these steps:

1. Replace z with x + yi.
2. Use the definition of the conjugate of a complex number: $\overline{z} = x - yi$.
3. Substitute z and $\overline{z}$ into the equation $|z+\overline{z}+1-i|=2$.
4. Simplify the equation and isolate the real and imaginary parts.
5. Express the locus of points in terms of x and y.

Here&#x27;s a detailed explanation using Latex:

We start by replacing z with x + yi:
$$ |(x+yi) + (x-yi) + 1 - i| = 2 $$

Simplify the expression inside the absolute value:
$$ |2x + 1 - i| = 2 $$

Now, let&#x27;s express 2x + 1 - i in terms of its real and imaginary parts:
$$ 2x + 1 - i = a + bi $$
where a is the real part and b is the imaginary part.

Squaring both sides to get rid of the absolute value:
$$ (a+bi)(a-bi) = 4 $$

Expanding and simplifying:
$$ a^2 - b^2 + 2abi = 4 $$

Comparing real and imaginary parts:
$$ a^2 - b^2 = 4 $$
$$ ab = 0 $$

Solving for a in terms of b from ab=0 gives us two cases: either a=0 or b=0.

Case 1: When a=0, we have
$$ -b^2 = 4 $$
which gives
$$ b^2 = -4 $$
This has no real solutions, so we discard this case.

Case 2: When b=0, we have
$$ a^2 = 4 $$
which gives
$$ a = ±\sqrt{4} $$
So, 
$$ a = ±2 $$

Therefore, combining both cases, we get two lines as loci:
When $a=+2$, it represents one line,
and when $a=-2$, it represents another line.

Finally, expressing these lines in terms of x and y gives us:
The locus of points satisfying $|z+\overline{z}+1-i| = 2$ is:
$y=\pm\sqrt{x^2+x+\frac{1}{4}}-1$