practice imaginary numbers

Simplify $\frac{1-i}{1+i}$.

$=\frac{1-i}{1+i}\cdot \frac{1-i}{1-i}$

$=\frac{1-2i-1}{1^2-i^2}$

$=\frac{-2i}{1+1}$

$=\frac{-2i}{2}$

$=-i$

Simplify $\frac{(1-i{)}^9}{(1+i{)}^8}$

$$\begin{array}{rl}\frac{(1-i{)}^9}{(1+i{)}^8}& =\frac{(1-i{)}^8(1-i)}{(1+i{)}^8}\\ & ={\left[\frac{(1-i)}{(1+i)}\right]}^8(1-i)\\ & ={\left[\frac{(1-i)(1-i)}{(1+i)(1-i)}\right]}^8(1-i)\\ & ={\left[\frac{1-2i+i^2}{1-i^2}\right]}^8(1-i)\\ & ={\left[\frac{1-2i-1}{1-(-1)}\right]}^8(1-i)\\ & ={\left[\frac{-2i}{2}\right]}^8(1-i)\\ & =(-i{)}^8(1-i)\\ & =((-1)i{)}^8(1-i)\\ & =((-1{)}^8{i}^8)(1-i)\\ & =(1\cdot (i^4{)}^2)(1-i)\\ & =(1\cdot (1{)}^2)(1-i)\\ & =1\cdot (1-i)\\ & =1-i\end{array}$$