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I started learn complex numbers and solve complex numbers. I tried to do this example but I don't really know and don't understand how to do it.
I know that $z = a + bi$
$a= \text{Re}$ and $b=\text{Im}$

Determine the complex number $$ if $_1= 2 + $ and is valid: $$\text{Re}(\frac{z}{z_{1}})=-\frac{3}{5},$$ and

$$\text{Im} (\bar{z}z_{1}) = 1$$

Américo Tavares
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community May 08 '22 at 21:58
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    These conditions can be rewritten as giving the real and imaginary parts, and hence value as a complex number, of $z\bar{z}_1$; then you need only divide by $\bar{z}_1$. – J.G. May 08 '22 at 22:00
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    Hint: if $z=a+ib$ then $\frac{1}{z}=\frac{a}{a^2+b^2}+i\frac{-b}{a^2+b^2}$ you can prove this multiplying $\frac{1}{z}*\frac{\bar{z}}{\bar{z}}$

    and $z\bar{z}=a^2-b^2$

    Use this to make a system of equations

     – aangulog May 08 '22 at 22:55
  • Hint: you have $,\text{Re}\left(\frac{z}{z_1}\right)=-\frac{3}{5},$ and $,\text{Im}\left(\frac{z}{z_1}\right)=-\frac{1}{|z_1|^2},$ so you can easily find $,\frac{z}{z_1},$. – dxiv May 08 '22 at 23:12

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There are many ways one could carry out the calculations to "extract" $ \ z \ $ from the given equations, as suggested in the comments above. A helpful identity for a beginner to know is that for $ \ z \ = \ a + bi \ \ , $ $ z·\overline{z} \ = \ |z|^2 \ = \ a^2 + b^2 \ \ . \ $ With $ \ z_1 \ = \ 2 + i \ \Rightarrow \ z_1·\overline{z_1} \ = \ |z_1|^2 \ = \ 2^2 + 1^2 \ = \ 5 \ \ , \ \mathfrak{Re} (\frac{z}{z_{1}}) \ = \ -\frac35 \ , \ \ $ and $ \ \mathfrak{Im} (\bar{z}z_{1}) \ = \ 1 \ \ , $ it may be also helpful to get the same product of factors to appear in the expressions we are working with. So we can take $$ \frac{z}{z_1} \ · \ \frac{\overline{z_1}}{\overline{z_1}} \ \ = \ \ \frac{z \ · \ \overline{z_1}}{z_1\ · \ \overline{z_1}} \ \ = \ \ \frac{z \ · \ \overline{z_1}}{5} \ \ \ \text{and} \ \ \ \overline{\overline{z} \ · \ z_1} \ \ = \ \ z \ · \ \overline{z_1} \ \ . $$

We then have $$ \mathfrak{Re} \left(\frac{z}{z_{1}} \right) \ \ = \ \ \mathfrak{Re} \left( \frac{z \ · \ \overline{z_1}}{5} \right) \ \ = \ -\frac35 \ \ \Rightarrow \ \ \mathfrak{Re} ( z \ · \ \overline{z_1}) \ \ = \ -3 \ \ \ , $$ $$ \mathfrak{Im} (\bar{z} \ · \ z_{1}) \ \ = \ \ 1 \ \ \Rightarrow \ \ \mathfrak{Im} (\overline{\overline{z} \ · \ z_1}) \ \ = \ \ -1 \ \ = \ \ \mathfrak{Im} (z \ · \ \overline{z_1}) \ \ . $$

Thus, we have found that $ \ z \ · \ \overline{z_1} \ = \ -3 - i \ \ . \ $ Upon multiplying this equation through by $ \ z_1 \ \ , \ $ we obtain $$ z \ · \ (\overline{z_1} · z_1) \ \ = \ \ (-3 \ - \ i) \ · \ z_1 \ \ = \ \ (-3 \ - \ i) \ · \ (2 \ + \ i) \ \ $$ $$ \Rightarrow \ \ z \ · \ 5 \ \ = \ \ -6 \ - \ 3i \ - \ 2i \ - \ i^2 \ \ = \ \ (-6 + 1) + (-3 - 2)·i \ \ = \ \ -5 \ - \ 5i $$ $$ \Rightarrow \ \ \boxed{ \ z \ \ = \ \ -1 \ - \ i \ } \ \ . $$

[Computing $ \ \frac{z}{z_{1}} \ = \ \frac{-1 \ - \ i}{2 \ + \ i} \ = \ \frac{(-1 \ - \ i)·(2 \ - \ i)}{5} \ = \ \frac{-2 \ - \ 2i \ + \ i \ - \ 1}{5} \ = \ \boxed{ \ -\frac35 \ } - \frac15·i \ \ $ and $ \ \bar{z}·z_{1} \ = \ (-1 + i)·(2 + i) \ = \ -2 \ - \ i \ + \ 2i \ - \ 1 \ = \ -3 \ + \ \boxed{ \ 1 \ } · i \ $ confirms that the specified conditions are satisfied.]