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. 

 A: The essential reason why your solution does not work is because $b$ is not necessarily a real number--in fact, your method of solution actually proves it cannot be real.  Therefore, your solution breaks down whenever you attempt to equate real and imaginary parts on each side of an equation.  Let's take the cross-multiplied equation $$20 + 10bi = (1-bi)(7+9i) = (7 + 9b) + (9 - 7b)i.$$  Now, if $b$ were real, this leads to an inconsistent solution as you have already observed.  So alternately, we collect like terms in $b$ to obtain $$0 = (-13 + 9i) + (9 - 17i)b,$$ or $$b = \frac{13 - 9i}{9 - 17i} = \frac{(13 - 9i)(9 + 17i)}{9^2 + 17^2} = \frac{27 + 14i}{37}.$$
A: 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 $$\frac{3b}{1+b^2}=\frac{9}{10}$$
A: You must have calculated wrongly. You should get $\frac{13-9i}{9-17i}$ by simplifying $20 + 10bi = (7+9i)(1-bi) = 7 - 7bi + 9i + 9b$.
