Answer
$0$.
Work Step by Step
The given function is
$\Rightarrow f(x)=x^2-2x+5$.
Replace $x$ with $1-2i$.
$\Rightarrow f(1-2i)=(1-2i)^2-2(1-2i)+5$.
Use the distributive property and special formula $(A-B)^2=A^2-2AB+B^2$.
$\Rightarrow f(1-2i)=1^2-2(1)(2i)+(2i)^2-2(1)+2(2i)+5$.
Simplify.
$\Rightarrow f(1-2i)=1-4i+4i^2-2+4i+5$.
Use $i^2=-1$.
$\Rightarrow f(1-2i)=1-4i-4-2+4i+5$.
Add like terms.
$\Rightarrow f(1-2i)=0$.
