Answer:
If f(x) = 1 – x then, value of |f(i)| is:
√2
Step-by-step explanation:
f(x) = 1 – x
Then, f(i) = 1 - i
We know that if f(z)=x+iy then, |f(z)|=\(\sqrt{x^2+y^2}\)
Hence, |f(i)| = \(\sqrt{1^2+(-1)^2}\)
= \(\sqrt{1+1}\)
= √2
Hence, if f(x) = 1 – x then, value of |f(i)| is:
√2
