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if $|z|-z=3-i\sqrt{3}$ determine the value of $|z|$

Below is the solution that I have, but I don't know how they get the (2) form (1) $$|z|-z=3-i\sqrt{3} \Longleftrightarrow z=|z|-3+i\sqrt{3}\;\; (1)$$ $$ \Rightarrow |z|^2=(|z|-3)^2+\sqrt{3}^2 \;\; (2)$$ $$\Leftrightarrow |z|^2=|z|^2-6|z|+12$$ $$\Leftrightarrow |z|=\frac{12}{6}=2$$

Ted Shifrin
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user579102
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    Equation (1) tells you the real and complex parts of $z$. Equation (2) is just the usual formula for $|z|^2 = (\Re z)^2+(\Im z)^2$. – Chris Lewis Jul 04 '23 at 20:29

3 Answers3

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Well, notice that when $\text{z}=\alpha+\beta i$ (with $\alpha\in\mathbb{R}$ and $\beta\in\mathbb{R}$) we get:

$$\left|\text{z}\right|-\text{z}=\sqrt{\alpha^2+\beta^2}-\alpha-\beta i\tag1$$

So, we can see:

  • $$\Re\left(\left|\text{z}\right|-\text{z}\right)=\sqrt{\alpha^2+\beta^2}-\alpha\tag2$$
  • $$\Im\left(\left|\text{z}\right|-\text{z}\right)=-\beta\tag3$$

So, we need to solve:

$$ \begin{cases} \begin{aligned} \sqrt{\alpha^2+\beta^2}-\alpha&=3\\ \\ -\beta&=-\sqrt{3} \end{aligned} \end{cases}\space\Longleftrightarrow\space \begin{cases} \begin{aligned} \alpha&=-1\\ \\ \beta&=\sqrt{3} \end{aligned} \end{cases}\tag4 $$

So, we end up with:

$$\left|\text{z}\right|=\sqrt{\left(-1\right)^2+\left(\sqrt{3}\right)^2}=\sqrt{4}=2\tag5$$

Jan Eerland
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They took the absolute value and squared it. Equation (1) splits $z$ up into its real part of $|z| - 3$ and its imaginary part of $\sqrt3$. From this, you can calculate the absolute value of $z$.

Gollol
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We have that

$$z=|z|-3+i\sqrt{3} =x+iy\implies x=\left|z\right|-3 \;\land\; y=\sqrt 3$$

therefore since $\left|z\right|=\sqrt{x^2+y^2}$

$$x=\sqrt{x^2+3}-3 \iff x=-1$$

and $\left|z\right|=x+3=2$.

user
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