<s>[INST] Finding $b$ such that $\frac{2+bi}{1-bi}=\frac7{10}+\frac9{10}i$. Where Have I Gone Wrong? 
Find $b$ where 
  $$\frac{2+bi}{1-bi}=\frac7{10}+\frac9{10}i$$

What I did to solve this problem was to rewrite the problem as 
$$\frac{2+bi}{1-bi}=\frac{7+9i}{10} \tag{1}$$ 
Given this, I then cross multiplied 
$$(2+bi)\cdot 10=(1-bi)(7+9i) \tag{2}$$
Then to 
$$20+10bi=(7+9b)+(9-7b)i \tag{3}$$
Then I got 
$$7+9b=20 \quad\text{and}\quad 9-7b=10b \tag{4}$$ 

The solutions I got from solving these where $13/9$ and $9/17$.
This was not the correct solution, however. Can someone please show me where I have gone wrong? I don't just want an alternative solution. I'm hoping for a detailed explanation on why my method doesn't work. Thank you. 

 [/INST] We have $$\frac{2+bi}{1-bi}=\frac{(2+bi)(1+bi)}{1+b^2}=\frac{2-b^2+3bi}{1+b^2}=\frac{2-b^2}{1+b^2}+\frac{3bi}{1+b^2}$$
and you will get
$$\frac{2-b^2}{1+b^2}=\frac{7}{10}$$ and $$\frac{3b}{1+b^2}=\frac{9}{10}$$
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