solving complex roots for a particular equation

Question

I am reading this book. I don't understand how did he get the second answer $1 + i$.

Solve the quadratic equation $z^2 - 3z + 3 + i = 0$.

Answer 1

The roots are$$\frac{3+(1-2i)}2\quad\text{and}\quad\frac{3-(1-2i)}2$$and$$\frac{3+(1-2i)}2=2-i\quad\text{and}\quad\frac{3-(1-2i)}2=1+i.$$

Answer 2

You know that $\sqrt{-3-4i} = 1-2i$ and that $z = \frac{3\pm\sqrt{-3-4i}}{2}$.

All you then have to do is plug in the value of the square root, i.e.

$$z = \frac{3\pm1-2i}{2}$$ $$z_{1}=\frac{3+1-2i}{2}=2-i$$

$$z_{1}=\frac{3-(1-2i)}{2}=\frac{3-1+2i)}{2}=1+i$$

Hope this answers your question.

Answer 3

$$z^2 - 3z + 3 + i = 0\to \\ z=\frac{3\pm\sqrt{9-4(1)(3+i)}}{2}\\ z=\frac{3\pm\sqrt{-3-4i}}{2}\\ z=\frac{3\pm\sqrt{1-4-2(2i)}}{2}\\ z=\frac{3\pm\sqrt{(1-2i)^2}}{2}$$ take a look at $$(1-2i)^2=1^2+4i^2-4i=1-4-4i$$ so $$z=\frac{3\pm\sqrt{(1-2i)^2}}{2}=\frac{3\pm(1-2i)}{2}$$