How can I get $1-2e^{-jw} + e^{-2jw}=(1-e^{-jw})^2$?

Question

how do you factor equations like these:

factor this: $1-2e^{-jw} + e^{-2jw}$ to this: $(1-e^{-jw})^2$ ?

Answer 1

Hint : $a^2-2ab+b^2=(a-b)^2$

Here $a=1$ and $b=e^{-jw}$

Therefore ; $$(1-e^{iw})^2=1^2+(e^{-iw})^2-2 \times 1\times e^{iw}=1 + e^{-2jw}-2e^{-jw}$$

Answer 2

The example you have given is just the binomial formula $(a - b)^2 = a^2 - 2ab + b^2$.