|z-1|+z=3-i, z= (7/4) - i

I can't find the original question, here is the answer for

$|z-1|+z=3-i$

1. Formula:

$\begin{array}{|rcll|} \hline z &=& a+bi \\ |z| &=& \sqrt{a^2+b^2} \\ \hline \end{array}$

2. |z-1|:

$\begin{array}{|rcll|} \hline |z-1| = |a+bi-1| &=& |(a-1)+bi| \\ &=& \sqrt{(a-1)^2+b^2} \\ &=& \sqrt{a^2-2a+1+b^2} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline |z-1|+z &=& 3-i \\\\ \overbrace{\underbrace{\sqrt{a^2-2a+1+b^2} +a}_{=3}}^{Re(z)}+\overbrace{\underbrace{b}_{=-1}\cdot i}^{Im(z)} &=& 3-1\cdot i \\\\ \Rightarrow \mathbf{b} & \mathbf{=} & \mathbf{-1} \\\\ \Rightarrow \sqrt{a^2-2a+1+b^2} +a &=&3 \quad &| \quad b=-1\\ \sqrt{a^2-2a+1+(-1)^2} +a &=& 3 \\ \sqrt{a^2-2a+1+1} +a &=& 3 \\ \sqrt{a^2-2a+2} +a &=& 3 \quad &| \quad - a\\ \sqrt{a^2-2a+2} &=& 3-a \quad &| \quad \text{square both sides} \\ a^2-2a+2 &=& (3-a)^2 \\ \not{a^2}-2a+2 &=& 9-6a+\not{a^2} \\ -2a+2 &=& 9-6a \quad &| \quad +6 a\\ 6a-2a+2 &=& 9 \quad &| \quad -2 \\ 4a &=& 9-2 \\ 4a &=& 7 \quad &| \quad :4\\ \mathbf{a} & \mathbf{=} & \mathbf{\frac74} \\ \hline \end{array}$

3. z:

$\begin{array}{|rcll|} \hline z &=& a+bi \\ z &=& \frac74-1\cdot i \\ \mathbf{z} & \mathbf{=} & \mathbf{\frac74- i} \\ \hline \end{array}$